The Bhargava Greedoid As a Gaussian Elimination Greedoid

The Bhargava greedoid as a Gaussian elimination greedoid

Darij Grinberg* June 10, 2021

Abstract. Inspired by Manjul Bhargava’s theory of generalized factorials, Fedor Petrov and the author have defined the Bhargava greedoid – a greedoid (a matroid-like set system on a finite set) assigned to any “ultra triple” (a somewhat extended variant of a finite ultrametric space). Here we show that the Bhargava greedoid of a finite ultra triple is always a Gaussian elimination greedoid over any sufficiently large (e.g., infinite) field; this is a greedoid analogue of a representable matroid. We find necessary and sufficient conditions on the size of the field to ensure this.

Contents

1. Gaussian elimination greedoids4 1.1. The deﬁnition ...... 4 1.2. Context ...... 5

2. V-ultra triples6

3. The main theorem9

4. Cliques and stronger bounds 10

5. The converse direction 11

6. Valadic V-ultra triples 12

7. Isomorphism 25

8. Decomposing a V-ultra triple 26

*Drexel University, Korman Center, Room 291, 15 S 33rd Street, Philadelphia PA, 19104, USA

1 The Bhargava greedoid as a Gaussian elimination greedoid page 2

9. Valadic representation of V-ultra triples 30

10.Proof of the main theorem 36

11.Proof of Theorem 5.1 36 11.1. Closed balls ...... 36 11.2. Exchange results for sets intersecting a ball ...... 39 11.3. Gaussian elimination greedoids in terms of determinants ...... 42 11.4. Proving the theorem ...... 48

12.Appendix: Gaussian elimination greedoids are strong 51

13.Appendix: Proof of Proposition 1.7 55

14.Appendix: Proofs of some properties of L 58

***

The notion of a greedoid was coined in 1981 by Korte and Lóvász, and has since seen significant developments ([KoLoSc91], [BjoZie92]). It is a type of set system (i.e., a set of subsets of a given ground set) that is required to satisfy some axioms weaker than the matroid axioms – so that, in particular, the independent sets of a matroid form a greedoid. In [GriPet19], Fedor Petrov and the author have constructed a greedoid stem- ming from Bhargava’s theory of generalized factorials, albeit in a setting signifi- cantly more general than Bhargava’s. Roughly speaking, the sets that belong to this greedoid are subsets of maximum perimeter (among all subsets of their size) of a finite ultrametric space. More precisely, our setup is more general than that of an ultrametric space: We work with a finite set E, a distance function d that assigns a “distance” d (e, f ) to any pair (e, f ) of distinct elements of E, and a weight function w that assigns a “weight” w (e) to each e ∈ E. The distances and weights are supposed to belong to a totally ordered abelian group V (for example, R). The distances are required to satisfy the symmetry axiom d (e, f ) = d ( f , e) and the “ultrametric triangle inequality” d (a, b) 6 max {d (a, c) , d (b, c)}. In this setting, any subset S of E has a well-defined perimeter, obtained by summing the weights and the pairwise distances of all its elements. The subsets S of E that have maximum perimeter (among all |S|-element subsets of E) then form a greedoid, which has been called the Bhargava greedoid of (E, w, d) in [GriPet19]. This greedoid is furthermore a strong greedoid [GriPet19, Theorem 6.1], which implies in particular that for any given k 6 |E|, the k-element subsets of E that have maximum perimeter are the bases of a matroid. In the present paper, we prove that the Bhargava greedoid of (E, w, d) is a Gaus- sian elimination greedoid over any sufficiently large (e.g., infinite) field (a greedoid The Bhargava greedoid as a Gaussian elimination greedoid page 3 analogue of a representable matroid1). We quantify the “sufficiently large” by pro- viding a sufficient condition for the size of the field. When all weights w (e) are equal, we show that this condition is also necessary. We note that the Bhargava greedoid can be seen to arise from an optimization problem in phylogenetics: Given a finite set E of organisms and an integer k ∈ N, we want to choose a k-element subset of E that maximizes some kind of biodiversity. Depending on the definition of biodiversity used, the properties of the maximizing subsets can differ. It appears natural to define biodiversity in terms of distances on the evolutionary tree (which is a finite ultrametric space), and such a definition has been considered by Moulton, Semple and Steel in [MoSeSt06], leading to the result that the maximum-biodiversity sets form a strong greedoid. The Bhargava greedoid is an analogue of their greedoid using a slightly different definition of biodiversity2. The present paper potentially breaks this analogy by showing that the Bhargava greedoid is a Gaussian elimination greedoid, whereas this is unknown for the greedoid of Moulton, Semple and Steel. Whether the latter is a Gaussian elimination greedoid as well remains to be understood3, as does the question of interpolating between the two notions of biodiversity.

This paper is self-contained (up to some elementary linear algebra), and in particular can be read independently of [GriPet19]. The 12-page extended abstract [GriPet20] summarizes the highlights of both [GriPet19] and this paper; it is thus a convenient starting point for a reader in- terested in the subject.

Acknowledgments I thank Fedor Petrov for his (major) part in the preceding project [GriPet19] that led to this one.

1In particular, this entails that all the matroids mentioned in the preceding paragraph are representable. 2To be specific: We view the organisms as the leaves of an evolutionary tree T that obeys a molecular clock assumption (i.e., all its leaves have the same distance from the root). Then, the set E of these organisms is equipped with a distance function (measuring distances along the edges of the tree), which satisfies the “ultrametric triangle inequality”. We define the weight function w by setting w (e) = 0 for all e ∈ E. Now, the phylogenetic diversity of a subset S ⊆ E is defined to be the sum of the edge lengths of the minimal subtree of T that connects all leaves in S. This phylogenetic diversity is the measure of biodiversity used in [MoSeSt06]. Meanwhile, our notion of perimeter can also be seen as a measure of biodiversity – perhaps even a better one for sustainability questions, as it rewards subsets that are roughly balanced across different clades. To give a trivial example, a zoo optimized using phylogenetic diversity might have dozens of mammals and only one bird, while this would unlikely be considered optimal in terms of perimeter. The molecular clock assumption can actually be dropped, at the expense of changing the weight function to account for different distances from the root. 3This question might have algorithmic significance. At least for polymatroids, representability can make the difference between a problem being NP-hard and in P, as shown by Lóvasz in [Lovasz80] for polymatroid matching. The Bhargava greedoid as a Gaussian elimination greedoid page 4

This work has been started during a Leibniz fellowship at the Mathematisches Forschungsinstitut Oberwolfach, and completed at the Institut Mittag-Lefﬂer in Djursholm; I thank both institutes for their hospitality. This material is based upon work supported by the Swedish Research Council under grant no. 2016-06596 while the author was in residence at Institut Mittag-Lefﬂer in Djur- sholm, Sweden during Spring 2020.

1. Gaussian elimination greedoids

1.1. The deﬁnition Convention 1.1. Here and in the following, N denotes the set {0, 1, 2, . . .}.

Convention 1.2. If E is any set, then 2E will denote the powerset of E (that is, the set of all subsets of E).

Convention 1.3. Let K be any field, and let n ∈ N. Then, Kn shall denote the K-vector space of all column vectors of size n over K. We recall the definition of a Gaussian elimination greedoid: Definition 1.4. Let E be a finite set. Let m ∈ N be such that m > |E|. Let K be a field. For each k ∈ {0, 1, . . . , m}, let π : Km → Kk be the projection map that removes all but the first k coordinates k       a1 a1 a1  a   a   a   2   2   2  m of a column vector. (That is, πk  .  =  .  for each  .  ∈ K .)  .   .   .  am ak am m For each e ∈ E, let ve ∈ K be a column vector. The family (ve)e∈E will be called a vector family over K. Let G be the subset

F |F| F ⊆ E | the family π|F| (ve) ∈ K is linearly independent e∈F

of 2E. Then, G is called the Gaussian elimination greedoid of the vector family (ve)e∈E. It is furthermore called a Gaussian elimination greedoid on ground set E.

Example 1.5. Let K = Q and E = {1, 2, 3, 4, 5} and m = 6. Let v1, v2, v3, v4, v5 ∈ K6 be the columns of the 6 × 5-matrix  0 1 1 0 1   1 1 0 0 0     0 2 1 0 1    .  1 0 1 0 0     0 0 0 0 0  1 2 0 2 1 The Bhargava greedoid as a Gaussian elimination greedoid page 5

Then, the Gaussian elimination greedoid of the vector family (ve)e∈E = (v1, v2, v3, v4, v5) is the set

{∅, {2} , {3} , {5} , {1, 2} , {1, 3} , {1, 5} , {2, 3} , {2, 5} , {1, 2, 3} , {1, 2, 5} , {1, 2, 3, 5}} .

For example, the 3-element set {1, 2, 5} belongs to this greedoid because the 3{1,2,5} family (π3 (ve))e∈{1,2,5} ∈ K is linearly independent (indeed, this family  0   1   1  consists of the vectors  1 ,  1  and  0 ). 0 2 1

Our definition of a Gaussian elimination greedoid follows [Knapp18, §1.3], ex- cept that we are using vector families instead of matrices (but this is equivalent, since any matrix can be identified with the vector family consisting of its columns) and we are talking about linear independence rather than non-singularity of matrices (but this is again equivalent, since a square matrix is non-singular if and only if its columns are linearly independent). The same definition is given in [KoLoSc91, §IV.2.3].

1.2. Context In the rest of Section 1, we shall briefly connect Definition 1.4 with known concepts in the theory of greedoids. This is not necessary for the rest of our work, so the impatient reader can well skip to Section 2. As the name suggests, Gaussian elimination greedoids are instances of greedoids – a class of set systems (i.e., sets of sets) characterized by some simple axioms. We refer to Definition 12.1 below for the definition of a greedoid, and to [KoLoSc91] for the properties of such. A subclass of greedoids that has particular interest to us are the strong greedoids; see, e.g., Section 12 below or [GriPet19, §6.1] or [BrySha99, §2] for their definition.4 The following theorem is implicit in [KoLoSc91, §IX.4]5:

Theorem 1.6. The Gaussian elimination greedoid G in Deﬁnition 1.4 is a strong greedoid.

See Section 12 below for a proof of this theorem. Matroids are a class of set systems more famous than greedoids; see [Oxley11] for their deﬁnition. We will not concern ourselves with matroids much in this note, but let us remark one connection to Gaussian elimination greedoids:6

4They also appear in [KoLoSc91, §IX.4] under the name of “Gauss greedoids”, but they are defined differently. (The equivalence between the two definitions is proved in [BrySha99, §2].) 5and partly proved in [Knapp18, §1.3] 6See [Oxley11, §1.1] for the definition of a representable matroid. The Bhargava greedoid as a Gaussian elimination greedoid page 6

Proposition 1.7. Let G be a Gaussian elimination greedoid on a ground set E. Let k ∈ N. Let Gk be the set of all k-element sets in G. Then, Gk is either empty or is the collection of bases of a representable matroid on the ground set E.

See Section 13 below for a proof of this proposition. Proposition 1.7 justiﬁes thinking of Gaussian elimination greedoids as a greedoid analogue of representable matroids.

2. V-ultra triples

Deﬁnition 2.1. Let E be a set. Then, E × E shall denote the subset {(e, f ) ∈ E × E | e 6= f } of E × E.

Convention 2.2. Fix a totally ordered abelian group (V, +, 0, 6) (with ground set V, group operation +, zero 0 and smaller-or-equal relation 6). The total order on V is supposed to be translation-invariant (i.e., if a, b, c ∈ V satisfy a 6 b, then a + c 6 b + c). We shall refer to the ordered abelian group (V, +, 0, 6) simply as V. We will use the standard additive notations for the abelian group V; in particular, we will use the ∑ sign for ﬁnite sums inside the group V. We will furthermore use the standard order-theoretical notations for the totally ordered set V; in particular, we will use the symbol > for the reverse relation of 6 (that is, a > b means b 6 a), and we will use the symbols < and > for the strict versions of the relations 6 and >. We will denote the largest element of a nonempty subset S of V (with respect to the relation 6) by max S. Likewise, min S will stand for the smallest element of S. For almost all examples we are aware of, it sufﬁces to set V to be the abelian group R, or even the smaller abelian group Z. Nevertheless, we shall work in full generality, as it serves to separate objects that would otherwise easily be confused.

Deﬁnition 2.3. A V-ultra triple shall mean a triple (E, w, d) consisting of:

• a set E, called the ground set of this V-ultra triple;

• a map w : E → V, called the weight function of this V-ultra triple;

• a map d : E × E → V, called the distance function of this V-ultra triple, and required to satisfy the following axioms: – Symmetry: We have d (a, b) = d (b, a) for any two distinct elements a and b of E. – Ultrametric triangle inequality: We have d (a, b) 6 max {d (a, c) , d (b, c)} for any three distinct elements a, b and c of E. The Bhargava greedoid as a Gaussian elimination greedoid page 7

If (E, w, d) is a V-ultra triple and e ∈ E, then the value w (e) ∈ V is called the weight of e. If (E, w, d) is a V-ultra triple and e and f are two distinct elements of E, then the value d (e, f ) ∈ V is called the distance between e and f .

Example 2.4. For this example, let V = Z, and let E be a subset of Z. Let m be any integer. Deﬁne a map w : E → V arbitrarily. Deﬁne a map d : E × E → V by ( 1, if a 6≡ b mod m; d (a, b) = for all (a, b) ∈ E × E. 0, if a ≡ b mod m

It is easy to see that (E, w, d) is a V-ultra triple.

The notion of a V-ultra triple generalizes the notion of an ultra triple as defined in [GriPet19]. More precisely, if V is the additive group (R, +, 0, 6) (with the usual addition and the usual total order on R), then a V-ultra triple is the same as what is called an “ultra triple” in [GriPet19]. It is straightforward to adapt all the definitions and results stated in [GriPet19] for ultra triples to the more general setting of V-ultra triples7. Let us specifically extend two definitions from [GriPet19] to V-ultra triples: the definition of a perimeter ([GriPet19, §3.1]) and the definition of the Bhargava greedoid ([GriPet19, §6.2]):

Definition 2.5. Let (E, w, d) be a V-ultra triple. Let A be a finite subset of E. Then, the perimeter of A (with respect to (E, w, d)) is defined to be ∑ w (a) + ∑ d (a, b) ∈ V. a∈A {a,b}⊆A; a6=b

(Here, the second sum ranges over all unordered pairs {a, b} of distinct elements of A.) The perimeter of A is denoted by PER (A).

For example, if A = {p, q, r} is a 3-element set, then

PER (A) = w (p) + w (q) + w (r) + d (p, q) + d (p, r) + d (q, r) .

Deﬁnition 2.6. Let S be any set, and let k ∈ N.A k-subset of S means a k-element subset of S (that is, a subset of S having size k).

7There is one stupid exception: The deﬁnition of R in [GriPet19, Remark 8.13] requires V 6= 0. But [GriPet19, Remark 8.13] is just a tangent without concrete use. The Bhargava greedoid as a Gaussian elimination greedoid page 8

Definition 2.7. Let (E, w, d) be a V-ultra triple such that E is finite. The Bhargava greedoid of (E, w, d) is defined to be the subset

{A ⊆ E | A has maximum perimeter among all |A| -subsets of E} = {A ⊆ E | PER (A) > PER (B) for all B ⊆ E satisfying |B| = |A|}

of 2E.

Example 2.8. For this example, let V = Z and E = {0, 1, 2, 3, 4}. Deﬁne a map w : E → V by setting w (e) = max {e, 1} for each e ∈ E. (Thus, w (0) = 1 and w (e) = e for all e > 0.) Deﬁne a map d : E × E → V by setting d (e, f ) = min {3, max {4 − e, 4 − f }} for all (e, f ) ∈ E × E. Here is a table of values of d: d 0 1 2 3 4 0 3 3 3 3 1 3 3 3 3 . 2 3 3 2 2 3 3 3 2 1 4 3 3 2 1 It is easy to see that (E, w, d) is a V-ultra triple. Let F be its Bhargava greedoid. Thus, F consists of the subsets A of E that have maximum perimeter among all |A|-subsets of E. What are these subsets?

• Clearly, ∅ is the only |∅|-subset of E, and thus has maximum perimeter among all |∅|-subsets of E. Hence, ∅ ∈ F. • The perimeter of a 1-subset {e} of E is just the weight w (e). Thus, the 1-subsets of E having maximum perimeter among all 1-subsets of E are precisely the subsets {e} where e ∈ E has maximum weight. In our example, there is only one e ∈ E having maximum weight, namely 4. Thus, the only 1-subset of E having maximum perimeter among all 1-subsets of E is {4}. In other words, the only 1-element set in F is {4}. • What about 2-element sets in F ? The perimeter PER {e, f } of a 2-subset {e, f } of E is w (e) + w ( f ) + d (e, f ). Thus, PER {0, 4} = w (0) + w (4) + d (0, 4) = 1 + 4 + 3 = 8 and similarly PER {1, 4} = 8 and PER {2, 4} = 8 and PER {3, 4} = 8 and PER {0, 3} = 7 and PER {1, 3} = 7 and PER {2, 3} = 7 and PER {0, 2} = 6 and PER {1, 2} = 6 and PER {0, 1} = 5. Thus, the 2-subsets of E having maximum perimeter among all 2-subsets of E are {0, 4} and {1, 4} and {2, 4} and {3, 4}. So these four sets are the 2-element sets in F. The Bhargava greedoid as a Gaussian elimination greedoid page 9

• Similarly, the 3-element sets in F are {0, 1, 4}, {0, 3, 4}, {1, 3, 4}, {0, 2, 4} and {1, 2, 4}. They have perimeter 15, while all other 3-subsets of E have perimeter 14 or 13.

• Similarly, the 4-element sets in F are {0, 1, 2, 4} and {0, 1, 3, 4}.

• Clearly, E is the only |E|-subset of E, and thus has maximum perimeter among all |E|-subsets of E. Hence, E ∈ F.

Thus, the Bhargava greedoid of (E, w, d) is

F = {∅, {4} , {0, 4} , {1, 4} , {2, 4} , {3, 4} , {0, 1, 4} , {0, 3, 4} , {1, 3, 4} , {0, 2, 4} , {1, 2, 4} , {0, 1, 2, 4} , {0, 1, 3, 4} , E} .

Example 2.9. For this example, let V = Z and E = {1, 2, 3}. Deﬁne a map w : E → V by setting w (e) = e for each e ∈ E. Deﬁne a map d : E × E → V by setting d (e, f ) = 1 for each (e, f ) ∈ E × E. It is easy to see that (E, w, d) is a V-ultra triple. Let F be the Bhargava greedoid of (E, w, d). What is F ? The same kind of reasoning as in Example 2.8 (but simpler due to the fact that all values of d are the same) shows that

F = {∅, {3} , {2, 3} , {1, 2, 3}} .

One thing we observed in both of these examples is the following simple fact:

Remark 2.10. Let (E, w, d) be a V-ultra triple such that E is ﬁnite. Let F be the Bhargava greedoid of (E, w, d). Then, E ∈ F.

Proof of Remark 2.10. The set E obviously has maximum perimeter among all |E|- subsets of E (since E is the only |E|-subset of E). But F is the Bhargava greedoid of (E, w, d). In other words, F = {A ⊆ E | A has maximum perimeter among all |A| -subsets of E} (by Deﬁnition 2.7). Hence, E ∈ F (since E is a subset of E that has maximum perimeter among all |E|-subsets of E). This proves Remark 2.10.

3. The main theorem

In [GriPet19, Theorem 6.1], it was proved that the Bhargava greedoid of an ultra triple with ﬁnite ground set is a strong greedoid8. More generally, this holds for any V-ultra triple with ﬁnite ground set (and the same argument can be used to prove this). However, we shall prove a stronger statement:

8See [GriPet19, §6.1] for the deﬁnition of strong greedoids. The Bhargava greedoid as a Gaussian elimination greedoid page 10

Theorem 3.1. Let (E, w, d) be a V-ultra triple such that E is ﬁnite. Let F be the Bhargava greedoid of (E, w, d). Let K be a ﬁeld of size |K| > |E|. Then, F is the Gaussian elimination greedoid of a vector family over K.

We will spend the next few sections working towards a proof of this theorem. First, however, let us extend it somewhat by strengthening the |K| > |E| bound.

4. Cliques and stronger bounds

For the rest of Section 3, we ﬁx a V-ultra triple (E, w, d). Let us deﬁne a certain kind of subsets of E, which we call cliques.

Deﬁnition 4.1. Let α ∈ V. An α-clique of (E, w, d) will mean a subset F of E such that any two distinct elements a, b ∈ F satisfy d (a, b) = α.

Deﬁnition 4.2. A clique of (E, w, d) will mean a subset of E that is an α-clique for some α ∈ V. Note that any 1-element subset of E is a clique (and an α-clique for every α ∈ V). The same holds for the empty subset. Any 2-element subset {a, b} of E is a clique and, in fact, a d (a, b)-clique. Note that the notion of a clique (and of an α-clique) depends only on E and d, not on w.

Example 4.3. For this example, let m, V, E, w and d be as in Example 2.4. Then: (a) The 0-cliques of E are the subsets of E whose elements are all mutually congruent modulo m. (b) The 1-cliques of E are the subsets of E that have no two distinct elements congruent to each other modulo m. Thus, any 1-clique has size 6 m if m is positive. (c) If α ∈ V is distinct from 0 and 1, then the α-cliques of E are the subsets of E having size 6 1. Using the notion of cliques, we can assign a number mcs (E, w, d) to our V-ultra triple (E, w, d):

Definition 4.4. Let mcs (E, w, d) denote the maximum size of a clique of (E, w, d). (This is well-defined whenever E is finite, and sometimes even otherwise.)

Clearly, mcs (E, w, d) 6 |E|, since any clique of (E, w, d) is a subset of E. Example 4.5. Let V, E, w and d be as in Example 2.8. Then, {0, 1, 2} is a 3- clique of (E, w, d) and has size 3; no larger cliques exist in (E, w, d). Thus, mcs (E, w, d) = 3. The Bhargava greedoid as a Gaussian elimination greedoid page 11

Example 4.6. For this example, let m, V, E, w and d be as in Example 2.4. Then: (a) If m = 2 and E = {1, 2, 3, 4, 5, 6}, then mcs (E, w, d) = 3, due to the 0-clique {1, 3, 5} having maximum size among all cliques. (b) If m = 3 and E = {1, 2, 3, 4, 5, 6}, then mcs (E, w, d) = 3, due to the 1-clique {1, 2, 3} having maximum size among all cliques.

We can now state a stronger version of Theorem 3.1:

Theorem 4.7. Let (E, w, d) be a V-ultra triple such that E is ﬁnite. Let F be the Bhargava greedoid of (E, w, d). Let K be a ﬁeld of size |K| > mcs (E, w, d). Then, F is the Gaussian elimination greedoid of a vector family over K.

Theorem 4.7 is stronger than Theorem 3.1 because |E| > mcs (E, w, d). We shall prove Theorem 4.7 in Section 10.

5. The converse direction

Before that, let us explore the question whether the bound |K| > mcs (E, w, d) can be improved. In an important particular case – namely, when the map w is constant9 –, it cannot, as the following theorem shows:

Theorem 5.1. Let (E, w, d) be a V-ultra triple such that E is ﬁnite. Assume that the map w is constant. Let F be the Bhargava greedoid of (E, w, d). Let K be a ﬁeld such that F is the Gaussian elimination greedoid of a vector family over K. Then, |K| > mcs (E, w, d).

We shall prove Theorem 5.1 in Section 11. When the map w in a V-ultra triple (E, w, d) is constant, Theorems 4.7 and 5.1 combined yield an exact characterization of those fields K for which the Bhargava greedoid of (E, w, d) can be represented as the Gaussian elimination greedoid of a vector family over K: Namely, those fields are precisely the fields K of size |K| > mcs (E, w, d). When w is not constant, Theorem 4.7 gives a sufficient condition; we don’t know a necessary condition. Here are two examples:

Example 5.2. Let V, E, w, d and F be as in Example 2.8. Then, mcs (E, w, d) = 3 (as we saw in Example 4.5). Hence, Theorem 4.7 shows that F can be represented as the Gaussian elimination greedoid of a vector family over any ﬁeld K of size |K| > 3. This bound on |K| is optimal, since the Bhargava greedoid F is not the Gaussian elimination greedoid of any vector family over the 2-element ﬁeld F2. (But this does not follow from Theorem 5.1, because w is not constant.)

9A map f : X → Y between two sets X and Y is said to be constant if all values of f are equal (i.e., if every x1, x2 ∈ X satisfy f (x1) = f (x2)). In particular, if |X| 6 1, then f : X → Y is automatically constant. The Bhargava greedoid as a Gaussian elimination greedoid page 12

Example 5.3. Let V, E, w, d and F be as in Example 2.9. Then, mcs (E, w, d) = 3, since E itself is a clique. Hence, Theorem 4.7 shows that F can be represented as the Gaussian elimination greedoid of a vector family over any ﬁeld K of size |K| > 3. However, this bound on |K| is not optimal. Indeed, the Bhargava greedoid F is the Gaussian elimination greedoid of the vector family (ve)e∈E =  0   0   1  (v1, v2, v3) over the ﬁeld F2, where v1 =  0 , v2 =  1  and v3 =  1 . 1 1 1

Question 5.4. Let (E, w, d) be a V-ultra triple such that E is finite. How to characterize the fields K for which the Bhargava greedoid of (E, w, d) is the Gaussian elimination greedoid of a vector family over K ? Is there a constant c (E, w, d) such that these fields are precisely the fields of size > c (E, w, d) ?

Remark 5.5. Let E, w, d and F be as in Theorem 3.1. Let K be any ﬁeld. For each k ∈ N, let Fk be the set of all k-element sets in F. If F is the Gaussian elimination greedoid of a vector family over K, then each Fk with k ∈ {0, 1, . . . , |E|} is the collection of bases of a representable matroid on the ground set E. (Indeed, this follows from Proposition 1.7, since Fk is nonempty.) But the converse is not true: It can happen that each Fk with k ∈ {0, 1, . . . , |E|} is the collection of bases of a representable matroid on the ground set E, yet F is not the Gaussian elimination greedoid of a vector family over K. For example, this happens if E = {1, 2, 3} and both maps w and d are constant E (so that F = 2 ), and K = F2.

6. Valadic V-ultra triples

As a first step towards the proof of Theorem 4.7, we will next introduce a special kind of V-ultra triples which, in a way, are similar to Bhargava’s for integers (see [GriPet19, Example 2.5 and §9]). We will call them valadic10, and we will see (in Theorem 6.9) that they satisfy Theorem 3.1. Afterwards (in Theorem 9.2), we will prove that any V-ultra triple with finite ground set is isomorphic (in an appropriate sense) to a valadic one over a sufficiently large field. Combining these two facts, we will then readily obtain Theorem 4.7.

10The name is a homage to the notion of a valuation ring, which is latent in the argument that follows (although never used explicitly). Indeed, if we deﬁne the notion of a valuation ring as in [Eisenb95, Exercise 11.1], then the K-algebra L+ constructed below is an instance of a valuation ring (with L being its fraction ﬁeld, and ord : L \ {0} → V being its valuation), and many of its properties that will be used below are instances of general properties of valuation rings. If we extended our argument to the more general setting of valuation rings, we would also recover Bhargava’s original ultra triples based on integer divisibility (see [GriPet19, Example 2.5 and §9]). However, we have no need for this generality (as we only need the construction as a stepping stone towards our proof of Theorem 4.7), and prefer to remain elementary and self-contained. The Bhargava greedoid as a Gaussian elimination greedoid page 13

Let us ﬁrst introduce some notations that will be used throughout Section 6.

Definition 6.1. We fix a field K. Let K [V] denote the group algebra of the group V over K. This is a free K-module with basis (tα)α∈V; it becomes a K-algebra with unity t0 and with multiplication determined by

tαtβ = tα+β for all α, β ∈ V.

This group algebra K [V] is commutative, since the group V is abelian. Let V>0 be the set of all α ∈ V satisfying α > 0; this is a submonoid of the group V. Let K [V>0] be the monoid algebra of this monoid V>0 over K. This is a K-algebra deﬁned in the same way as K [V], but using V>0 instead of V. It is clear that K [V>0] is the K-subalgebra of K [V] spanned by the basis elements tα with α ∈ V>0.

Example 6.2. If V = Z (with the usual addition and total order), then V>0 = N. In this case, the group algebra K [V] is the Laurent polynomial ring K X, X−1 in a single indeterminate X over K (indeed, t1 plays the role of X, and more gen- α erally, each tα plays the role of X ), and its subalgebra K [V>0] is the polynomial ring K [X].

Definition 6.3. (a) Let L be the commutative K-algebra K [V], and let L+ be its K-subalgebra K [V>0]. Thus, the K-module L has basis (tα)α∈V, while its K-submodule L+ has basis (tα)α∈V . >0 (b) If a ∈ L and β ∈ V, then tβ a shall denote the coefficient of tβ in a (when a is expanded in the basis (tα)α∈V of L). This is an element of K. For example, [t3](t2 − t3 + 5t6) = −1 (if V = Z). (c) If a ∈ L is nonzero, then the order of a is defined to be the smallest β ∈ V such that tβ a 6= 0. This order is an element of V, and is denoted by ord a. For example, ord (t2 − t3 + 5t6) = 2 (if V = Z). Note that ord (tα) = α for each α ∈ V.

Lemma 6.4. (a) A nonzero element a ∈ L belongs to L+ if and only if its order ord a is nonnegative (i.e., we have ord a > 0). (b) We have ord (−a) = ord a for any nonzero a ∈ L. (c) Let a and b be two nonzero elements of L. Then, ab is nonzero and satisﬁes ord (ab) = ord a + ord b. (d) Let a and b be two nonzero elements of L such that a + b is nonzero. Then, ord (a + b) > min {ord a, ord b}.

See Section 14 for the (straightforward) proof of this lemma.

Corollary 6.5. The ring L is an integral domain.

Proof of Corollary 6.5. This follows from Lemma 6.4 (c). The Bhargava greedoid as a Gaussian elimination greedoid page 14

Applying Lemma 6.4 (c) many times, we also obtain the following:

Corollary 6.6. The map ord : L \ {0} → V transforms (finite) products into sums. In more detail: If (ai)i∈I is any finite family of nonzero elements of L, then the product ∏ ai is nonzero and satisfies i∈I ! ord ∏ ai = ∑ ord (ai) . i∈I i∈I

Proof. Induction on |I|. The induction step uses Lemma 6.4 (c); the straightforward details are left to the reader. We can now assign a V-ultra triple to each subset of L:

Deﬁnition 6.7. Let E be a subset of L. Deﬁne a distance function d : E × E → V by setting d (a, b) = − ord (a − b) for all (a, b) ∈ E × E. (Recall that E × E means the set {(a, b) ∈ E × E | a 6= b}.) Then, (E, w, d) is a V-ultra triple whenever w : E → V is a function (by Lemma 6.8 below). Such a V-ultra triple (E, w, d) will be called valadic.

Lemma 6.8. In Deﬁnition 6.7, the triple (E, w, d) is indeed a V-ultra triple.

Lemma 6.8 follows easily from Lemma 6.4. (See Section 14 for the details of the proof.) Now, we claim that the Bhargava greedoid of a valadic V-ultra triple (E, w, d) with ﬁnite E is the Gaussian elimination greedoid of a vector family over K:

Theorem 6.9. Let E be a finite subset of L. Define d as in Definition 6.7. Let w : E → V be a function. Then, the Bhargava greedoid of the V-ultra triple (E, w, d) is the Gaussian elimination greedoid of a vector family over K.

In order to prove this theorem, we will need a determinantal identity:

Lemma 6.10. Let R be a commutative ring. Consider the polynomial ring R [X]. Let m ∈ N. Let f1, f2,..., fm be m polynomials in R [X]. Assume that fj is a monic polynomial of degree j − 1 for each j ∈ {1, 2, . . . , m}. Let u1, u2,..., um be m elements of R. Then, det fj (ui) = ui − uj . 16i6m, 16j6m ∏ (i,j)∈{1,2,...,m}2; i>j The Bhargava greedoid as a Gaussian elimination greedoid page 15

Here, we are using the notation bi,j for the p × q-matrix whose (i, j)-th 16i6p, 16j6q entry is bi,j for all i ∈ {1, 2, . . . , p} and all j ∈ {1, 2, . . . , q}. Lemma 6.10 is a classical generalization of the famous Vandermonde determinant. In this form, it is a particular case of [Grinbe11, Theorem 2] (applied to j−1 Pj = fj and ai = ui), because the coefﬁcient of X in a monic polynomial of degree j − 1 is 1. It also appears in [FadSom72, Exercise 267] (where it is stated for the transpose of the matrix we are considering here), in [Muir60, Chapter XI, Exercise 2 in Set XVIII] (where it, too, is stated for the transpose of the matrix), in [Kratte99, Proposition 1], and in [Grinbe15, Exercise 6.62]. We need two more simple lemmas for our proof of Theorem 6.9:

Lemma 6.11. The map

π : L+ → K,

x 7→ [t0] x

is a K-algebra homomorphism.

Lemma 6.12. Consider the map π : L+ → K from Lemma 6.11. Let a ∈ L+ be nonzero. Then, π (a) 6= 0 holds if and only if ord a = 0.

See Section 14 for the (easy) proofs of these two lemmas. Proof of Theorem 6.9. Let m = |E|. Consider the V-ultra triple (E, w, d); all perime- ters discussed in this proof are deﬁned with respect to this V-ultra triple. We construct a list (c1, c2,..., cm) of elements of E by the following recursive procedure:

• For each i ∈ {1, 2, . . . , m}, we choose ci (assuming that all the preceding entries c1, c2,..., ci−1 of our list are already constructed) to be an element of E \ {c1, c2,..., ci−1} that maximizes the perimeter PER {c1, c2,..., ci}. This procedure can indeed be carried out, since at each step we can find an element 11 ci ∈ E \ {c1, c2,..., ci−1} that maximizes the perimeter PER {c1, c2,..., ci}. Clearly, this procedure constructs an m-tuple (c1, c2,..., cm) of elements of E. The 12 m entries c1, c2,..., cm of this m-tuple are distinct , and thus are m distinct elements of E; but E has only m elements altogether (since m = |E|). Hence, the m entries c1, c2,..., cm must cover the whole set E. In other words, E = {c1, c2,..., cm}. Furthermore, for each i ∈ {1, 2, . . . , m} and each x ∈ E \ {c1, c2,..., ci−1}, we have PER {c1, c2,..., ci} > PER {c1, c2,..., ci−1, x} (1) 11 Indeed, the set E \ {c1, c2,..., ci−1} is nonempty (since |{c1, c2,..., ci−1}| 6 i − 1 < i 6 m = |E| and thus {c1, c2,..., ci−1} 6⊇ E) and finite (since E is finite), and thus at least one of its elements will maximize the perimeter in question. 12 since each ci is chosen to be an element of E \ {c1, c2,..., ci−1}, and thus is distinct from all the preceding entries c1, c2,..., ci−1 The Bhargava greedoid as a Gaussian elimination greedoid page 16

(due to how ci is chosen). Thus, in the parlance of [GriPet19, §3.2], the m-tuple (c1, c2,..., cm) is a greedy m-permutation of E. For each j ∈ {1, 2, . . . , m}, deﬁne a ρj ∈ V by

j−1 ρj = w cj + ∑ d ci, cj . (2) i=1 ◦ (This is precisely what is called νj (C) in [GriPet19], where C = E.) Consider the polynomial ring L [X]. For each j ∈ {1, 2, . . . , m}, deﬁne a polynomial fj ∈ L [X] by

j−1 fj = (X − c1)(X − c2) ··· X − cj−1 = ∏ (X − ci) . i=1 This is a monic polynomial of degree j − 1. Next we claim the following:

∈ ∈ { } ( ) ∈ L Claim 1: Let e E and j 1, 2, . . . , m . Then, tρj−w(e) fj e +.

j−1 j−1 [Proof of Claim 1: We have fj = ∏ (X − ci) and thus fj (e) = ∏ (e − ci). Hence, i=1 i=1 ∈ ( ) = ( ) ∈ L if e c1, c2,..., cj−1 , then fj e 0 and thus our claim tρj−w(e) fj e + is obvious. Thus, we WLOG assume that e ∈/ c1, c2,..., cj−1 . Thus, each i ∈ j−1 {1, 2, . . . , j − 1} satisﬁes e 6= ci and thus e − ci 6= 0. Hence, ∏ (e − ci) is a product i=1 of nonzero elements of L, and thus is itself nonzero (since Corollary 6.5 says that L j−1 is an integral domain). In other words, fj (e) is nonzero (since fj (e) = ∏ (e − ci)). i=1 ( ) L Hence, tρj−w(e) fj e is nonzero as well (since tρj−w(e) is nonzero, and since is an integral domain). j−1 Moreover, from fj (e) = ∏ (e − ci), we obtain i=1

j−1 ! j−1 ord fj (e) = ord ∏ (e − ci) = ∑ ord (e − ci) (3) i=1 i=1

(by Corollary 6.6). From e ∈ E and e ∈/ c1, c2,..., cj−1 , we obtain e ∈ E \ c1, c2,..., cj−1 . Hence, (1) (applied to i = j and x = e) yields PER c1, c2,..., cj > PER c1, c2,..., cj−1, e . (4) The Bhargava greedoid as a Gaussian elimination greedoid page 17

13 But c1, c2,..., cj are distinct . Hence, the deﬁnition of the perimeter yields

j PER c1, c2,..., cj = ∑ w (ci) + ∑ d ci, cp i=1 16i

14 (since c1, c2,..., cj−1, e are distinct ). Hence, (4) rewrites as

j−1 ρj + ∑ w (ci) + ∑ d ci, cp i=1 16i ∑ w (ci) + w (e) + ∑ d ci, cp + ∑ d (ci, e) . i=1 16i

j−1 ρj > w (e) + ∑ d (ci, e) . i=1 In view of j−1 j−1 j−1 ∑ d (ci, e) = ∑ d (e, ci) = − ∑ ord (e − ci) = − ord fj (e) , i=1 | {z } i=1 | {z } i=1 =d(e,ci) =− ord(e−ci) | {z } (by the “Symmetry” (by the deﬁnition of d) =ord( fj(e)) axiom in the deﬁnition of a V-ultra triple) (by (3))

13 This is because c1, c2,..., cm are distinct. 14 This is because c1, c2,..., cm are distinct and e ∈/ c1, c2,..., cj−1 . The Bhargava greedoid as a Gaussian elimination greedoid page 18 this rewrites as ρj > w (e) − ord fj (e) . In other words, ord fj (e) > w (e) − ρj. Now, Lemma 6.4 (c) (applied to a = = ( ) tρj−w(e) and b fj e ) yields ( ) = + ( ) ord tρj−w(e) fj e ord tρj−w(e) ord fj e | {z } | {z } w(e)−ρ =ρj−w(e) > j

> ρj − w (e) + w (e) − ρj = 0. = ( ) ( ) Hence, Lemma 6.4 (a) (applied to a tρj−w(e) fj e ) shows that tρj−w(e) fj e belongs L ( ) ∈ L to +. Thus, tρj−w(e) fj e +. This proves Claim 1.] For each e ∈ E and j ∈ {1, 2, . . . , m}, we deﬁne an a (e, j) ∈ L+ by ( ) = ( ) a e, j tρj−w(e) fj e . (5) (This is well-deﬁned, due to Claim 1.) We now claim the following:

Claim 2: Let k ∈ N. Let u1, u2,..., uk be any k distinct elements of E. Let U = {u1, u2,..., uk}. Then, det (a (u , j)) is a nonzero element of L (6) i 16j6k, 16i6k + and k ord det (a (u , j)) = ρ − PER (U) . (7) i 16j6k, 16i6k ∑ j j=1

[Proof of Claim 2: The set E has at least k many elements (since u1, u2,..., uk are k distinct elements of E). In other words, |E| > k. Hence, k 6 |E| = m. Therefore, {1, 2, . . . , k} ⊆ {1, 2, . . . , m}. In other words, for each j ∈ {1, 2, . . . , k}, we have j ∈ {1, 2, . . . , m}. Hence, a (ui, j) ∈ L+ for any i, j ∈ {1, 2, . . . , k} (since we deﬁned a (e, j) to satisfy a (e, j) ∈ L+ for any e ∈ E and j ∈ {1, 2, . . . , m}). In other words, all entries of the matrix (a (ui, j))1 j k, 1 i k belong to L+. Hence, its determinant 6 6 6 6 det (a (u , j)) belongs to L as well (since L is a ring). i 16j6k, 16i6k + + Lemma 6.10 (applied to L and k instead of R and m) yields det fj (ui) = ui − uj . (8) 16i6k, 16j6k ∏ (i,j)∈{1,2,...,k}2; i>j It is known that the determinant of a matrix equals the determinant of its transpose. Thus, det fj (ui) = det fj (ui) 16j6k, 16i6k 16i6k, 16j6k The Bhargava greedoid as a Gaussian elimination greedoid page 19

(since the matrix fj (ui) is the transpose of the matrix fj (ui) ). 16j6k, 16i6k 16i6k, 16j6k But when we scale a column of a matrix by a scalar λ, then its determinant also gets multiplied by λ. Hence,

k ! det t−w(u ) fj (ui) = t−w(u ) · det fj (ui) i 1 j k, 1 i k ∏ i 16j6k, 16i6k 6 6 6 6 i=1 | {z } =det ( fj(ui)) 16i6k, 16j6k = ∏ (ui−uj) (i,j)∈{1,2,...,k}2; i>j (by (8)) k ! = t · u − u . ∏ −w(ui) ∏ i j i=1 (i,j)∈{1,2,...,k}2; i>j

Furthermore, when we scale a row of a matrix by a scalar λ, then its determinant also gets multiplied by λ. Hence,

( ) det tρj t−w(ui) fj ui 16j6k, 16i6k ! k = tρ · det t−w(u ) fj (ui) ∏ j i 1 j k, 1 i k j=1 6 6 6 6 | {z } k = ∏ t · ∏ (ui−uj) −w(ui) i=1 (i,j)∈{1,2,...,k}2; i>j k ! k ! = t t · u − u . ∏ ρj ∏ −w(ui) ∏ i j j=1 i=1 (i,j)∈{1,2,...,k}2; i>j

But for every i ∈ {1, 2, . . . , k} and j ∈ {1, 2, . . . , k}, we have

( ) = ( )( ( )) a ui, j tρj−w(ui) fj ui by the deﬁnition of a ui, j | {z } =tρ t j −w(ui) = ( ) tρj t−w(ui) fj ui . The Bhargava greedoid as a Gaussian elimination greedoid page 20

Hence,         det  a (ui, j)    | {z }   = ( ) tρj t−w(u ) fj ui i 16j6k, 16i6k = ( ) det tρj t−w(ui) fj ui 16j6k, 16i6k k ! k ! = t t · u − u . (9) ∏ ρj ∏ −w(ui) ∏ i j j=1 i=1 (i,j)∈{1,2,...,k}2; i>j

The right hand side of this equality is a product of nonzero elements of L (since u , u2,..., u are distinct), and thus is nonzero (by Corollary 6.5). Hence, the left 1 k hand side is nonzero. In other words, det (a (u , j)) is nonzero. This i 16j6k, 16i6k proves (6) (since we already know that det (a (u , j)) belongs to L ). i 16j6k, 16i6k + Moreover, (9) yields ord det (a (u , j)) i 16j6k, 16i6k   ! !  k k  = ord  t t · u − u   ∏ ρj ∏ −w(ui) ∏ i j   j=1 i=1 (i,j)∈{1,2,...,k}2;  i>j k k = ord t + ord t + ord u − u ∑ ρj ∑ −w(ui) ∑ i j j=1 | {z } i=1 | {z } (i,j)∈{1,2,...,k}2; = ρj =−w(ui) i>j (by Lemma 6.4 (c) and Corollary 6.6) k k = ∑ ρj − ∑ w (ui) + ∑ ord ui − uj j=1 i=1 (i,j)∈{1,2,...,k}2; i>j   k  k    = ∑ ρj − ∑ w (ui) − ∑ ord ui − uj  . (10) j=1 i=1 (i,j)∈{1,2,...,k}2;  i>j

But recall that U = {u1, u2,..., uk} with u1, u2,..., uk distinct. The deﬁnition of The Bhargava greedoid as a Gaussian elimination greedoid page 21 perimeter thus yields

k PER (U) = ∑ w (ui) + ∑ d ui, uj i=1 16ij of a V-ultra triple) here, we have renamed the index (i, j) as (j, i) in the second sum k = ∑ w (ui) + ∑ d ui, uj i=1 (i,j)∈{1,2,...,k}2; | {z } i>j =− ord(ui−uj) (by the deﬁnition of d) k = ∑ w (ui) − ∑ ord ui − uj . (11) i=1 (i,j)∈{1,2,...,k}2; i>j

Hence, (10) becomes ord det (a (u , j)) i 16j6k, 16i6k   k  k    = ∑ ρj − ∑ w (ui) − ∑ ord ui − uj  j=1 i=1 (i,j)∈{1,2,...,k}2;  i>j | {z } =PER(U) (by (11)) k = ∑ ρj − PER (U) . j=1

Hence, (7) is proved. This proves Claim 2.] As a consequence of Claim 2, we obtain the following:

k Claim 3: Let k ∈ {0, 1, . . . , m}. Then, ∑ ρj is the maximum perimeter of j=1 a k-subset of E.

[Proof of Claim 3: The elements c1, c2,..., cm are distinct; thus, the elements c1, c2,..., ck are distinct. Hence, {c1, c2,..., ck} is a k-subset of E. The Bhargava greedoid as a Gaussian elimination greedoid page 22

Adding up the equalities (2) for all j ∈ {1, 2, . . . , k}, we obtain

k k j−1 ! ∑ ρj = ∑ w cj + ∑ d ci, cj j=1 j=1 i=1 k = ∑ w cj + ∑ d ci, cj = PER {c1, c2,..., ck} j=1 16i

(since c1, c2,..., ck are distinct). Since {c1, c2,..., ck} is a k-subset of E, we thus k conclude that ∑ ρj is the perimeter of some k-subset of E. Thus, in order to prove j=1 k Claim 3, we need only to show that ∑ ρj > PER (U) for every k-subset U of E. j=1 k So let U be a k-subset of E. We must prove ∑ ρj > PER (U). j=1 Write the k-subset U in the form U = {u1, u2,..., uk} for k distinct elements u1, u2,..., uk of E. Claim 2 thus yields that det (a (u , j)) is a nonzero element of L i 16j6k, 16i6k + and k ord det (a (u , j)) = ρ − PER (U) . (12) i 16j6k, 16i6k ∑ j j=1 Thus, in particular, det (a (u , j)) belongs to L . Hence, i 16j6k, 16i6k + ord det (a (u , j)) 0 i 16j6k, 16i6k > (by Lemma 6.4 (a), applied to det (a (u , j)) instead of a). In view of i 16j6k, 16i6k k k (12), this rewrites as ∑ ρj − PER (U) > 0. In other words, ∑ ρj > PER (U). This j=1 j=1 completes the proof of Claim 3.] Now, recall the K-algebra homomorphism π : L+ → K from Lemma 6.11. For m each e ∈ E, deﬁne a column vector ve ∈ K by  π (a (e, 1))   π (a (e, 2))  v =   = (π (a (e, j))) . e  .  16j6m  .  π (a (e, m))

We thus have a vector family (ve)e∈E over K. Let G be the Gaussian elimination greedoid of this family. Let F be the Bhargava greedoid of (E, w, d). Our goal is to prove that F = G (since this will yield Theorem 6.9). The Bhargava greedoid as a Gaussian elimination greedoid page 23

In order to do so, it sufﬁces to show that if U is any subset of E, then we have the logical equivalence (U ∈ F) ⇐⇒ (U ∈ G) . (13) So let us do this. Let U be a subset of E. We must prove the equivalence (13). Write the subset U in the form U = {u1, u2,..., uk} with u1, u2,..., uk distinct. Thus, |U| = k. In other words, U is a k-subset of E. Claim 2 thus yields that det (a (u , j)) is a nonzero element of L (14) i 16j6k, 16i6k + and k ord det (a (u , j)) = ρ − PER (U) . (15) i 16j6k, 16i6k ∑ j j=1 Also, recall that π is a K-algebra homomorphism (by Lemma 6.11), thus a ring homomorphism. Hence, det (π (a (u , j))) = π det (a (u , j)) (16) i 16j6k, 16i6k i 16j6k, 16i6k

(since ring homomorphisms respect determinants). Also, U is a subset of E; hence, |U| 6 |E| = m. Thus, k = |U| 6 m. Hence, k k ∈ {0, 1, . . . , m}. Therefore, Claim 3 shows that ∑ ρj is the maximum perimeter j=1 k of a k-subset of E. In other words, ∑ ρj is the maximum perimeter of a |U|-subset j=1 of E (since |U| = k). The deﬁnition of the Gaussian elimination greedoid G shows that we have the The Bhargava greedoid as a Gaussian elimination greedoid page 24 following equivalence:15

(U ∈ G) U |U| ⇐⇒ the family π|U| (ve) ∈ K is linearly independent e∈U U k ⇐⇒ the family (πk (ve))e∈U ∈ K is linearly independent (since |U| = k)

⇐⇒ (the vectors πk (vu1 ) , πk (vu2 ) ,..., πk (vuk ) are linearly independent) (since U = {u1, u2,..., uk} with u1, u2,..., uk distinct) ⇐⇒ the columns of the matrix (π (a (u , j))) are linearly independent i 16j6k, 16i6k ! since the vectors πk (vu1 ) , πk (vu2 ) ,..., πk (vuk ) are the columns of the matrix (π (a (u , j))) i 16j6k, 16i6k ⇐⇒ the matrix (π (a (u , j))) is invertible i 16j6k, 16i6k since a square matrix over the ﬁeld K is invertible if and only if its columns are linearly independent ⇐⇒ det (π (a (u , j))) 6= 0 in K i 16j6k, 16i6k ⇐⇒ π det (a (u , j)) 6= 0 in K (by (16)) i 16j6k, 16i6k ⇐⇒ ord det (a (u , j)) = 0 i 16j6k, 16i6k by Lemma 6.12, applied to det (a (u , j)) instead of a i 16j6k, 16i6k k ! ⇐⇒ ∑ ρj − PER (U) = 0 (by (15)) j=1 k ! ⇐⇒ PER (U) = ∑ ρj j=1 ⇐⇒ (PER (U) is the maximum perimeter of a |U| -subset of E) k ! since ∑ ρj is the maximum perimeter of a |U| -subset of E j=1 ⇐⇒ (U has maximum perimeter among all |U| -subsets of E) ⇐⇒ (U ∈ F)

(by the deﬁnition of the Bhargava greedoid F). Thus, the equivalence (13) is proven. This concludes the proof of Theorem 6.9.

15The words “linearly independent” should always be understood to mean “K-linearly independent” here. The Bhargava greedoid as a Gaussian elimination greedoid page 25

7. Isomorphism

Next, we introduce the notion of a set system. This elementary notion will play a purely technical role in what follows.

Deﬁnition 7.1. Let E be a set. (a) We let 2E denote the powerset of E. (b) A set system on ground set E shall mean a subset of 2E.

Thus:

• The Gaussian elimination greedoid of a vector family (ve)e∈E (over any ﬁeld K) is a set system on ground set E. • The Bhargava greedoid of any V-ultra triple (E, w, d) is a set system on ground set E.

(More generally, any greedoid is a set system, but we shall not need this.) We shall use the following two simple concepts of isomorphism:

Deﬁnition 7.2. Let (E, w, d) and (F, v, c) be two V-ultra triples. (a) A bijective map f : E → F is said to be an isomorphism of V-ultra triples from (E, w, d) to (F, v, c) if it satisﬁes v ◦ f = w and

c ( f (a) , f (b)) = d (a, b) for all (a, b) ∈ E × E.

(b) The V-ultra triples (E, w, d) and (F, v, c) are said to be isomorphic if there exists an isomorphism f : E → F of V-ultra triples from (E, w, d) to (F, v, c). (Note that being isomorphic is clearly a symmetric relation, because if f : E → F is an isomorphism of V-ultra triples from (E, w, d) to (F, v, c), then f −1 : F → E is an isomorphism of V-ultra triples from (F, v, c) to (E, w, d).)

Deﬁnition 7.3. Let E and F be two set systems on ground sets E and F, respectively. (a) A bijective map f : E → F is said to be an isomorphism of set systems from E to F if the bijection 2 f : 2E → 2F induced by it (i.e., the bijection that sends each S ∈ 2E to f (S) ∈ 2F) satisﬁes 2 f (E) = F. (b) The set systems E and F are said to be isomorphic if there exists an isomorphism f : E → F of set systems from E to F. (Note that being isomorphic is clearly a symmetric relation, because if f : E → F is an isomorphism of set systems from E to F, then f −1 : F → E is an isomorphism of set systems from F to E.)

The intuitive meaning of both of these two deﬁnitions is simple: Two V-ultra triples are isomorphic if and only if one can be obtained from the other by relabel- ing the elements of the ground set. The same holds for two set systems. The following two propositions are obvious: The Bhargava greedoid as a Gaussian elimination greedoid page 26

Proposition 7.4. Let (E, w, d) and (F, v, c) be two isomorphic V-ultra triples such that E and F are ﬁnite. Then, the Bhargava greedoids of (E, w, d) and (F, v, c) are isomorphic as set systems.

Proposition 7.5. Let K be a ﬁeld. Let E and F be two isomorphic set systems. If E is the Gaussian elimination greedoid of a vector family over K, then so is F.

8. Decomposing a V-ultra triple

For the rest of Section 8, we fix a V-ultra triple (E, w, d). We are going to study the structure of this V-ultra triple. We recall the notions of α-cliques and cliques (defined in Definition 4.1 and Definition 4.2, respectively). We shall next define another kind of subsets of E: the open balls.

◦ Definition 8.1. Let α ∈ V and e ∈ E. The open ball Bα (e) is defined to be the subset { f ∈ E | f = e or else d ( f , e) < α} of E. ◦ Clearly, for each α ∈ V and each e ∈ E, we have e ∈ Bα (e), so that the open ball ◦ Bα (e) contains at least the element e. Proposition 8.2. Let α ∈ V and e, f ∈ E be such that e 6= f and d (e, f ) < α. ◦ ◦ Then, Bα (e) = Bα ( f ). Proof. We have e 6= f , so that f 6= e. Hence, the “symmetry” axiom in Definition 2.3 yields that d ( f , e) = d (e, f ) < α. ◦ ◦ From e 6= f and d (e, f ) < α, we obtain e ∈ Bα ( f ) (by the definition of Bα ( f )). ◦ ◦ Also, f ∈ Bα ( f ) (by the definition of Bα ( f )). ◦ ◦ Let x ∈ Bα (e). We shall show that x ∈ Bα ( f ). ◦ If x = e, then this follows immediately from e ∈ Bα ( f ). Hence, for the rest of ◦ ◦ the proof of x ∈ Bα ( f ), we WLOG assume that x 6= e. Thus, from x ∈ Bα (e), ◦ ◦ we obtain d (x, e) < α (by the definition of Bα (e)). If x = f , then x ∈ Bα ( f ) ◦ follows immediately from the fact that f ∈ Bα ( f ). Thus, for the rest of the proof ◦ of x ∈ Bα ( f ), we WLOG assume that x 6= f . Now, the points e, f , x are distinct (since x 6= e, x 6= f and e 6= f ). Hence, the ultrametric triangle inequality yields d (x, f ) 6 max {d (x, e) , d ( f , e)} < α (since d (x, e) < α and d ( f , e) < α). Thus, ◦ ◦ x ∈ Bα ( f ) (by the definition of Bα ( f )). ◦ Now, forget that we fixed x. We thus have shown that x ∈ Bα ( f ) for each ◦ ◦ ◦ x ∈ Bα (e). In other words, Bα (e) ⊆ Bα ( f ). But our situation is symmetric in e and f (since f 6= e and d ( f , e) < α). Hence, ◦ ◦ the same argument that let us prove Bα (e) ⊆ Bα ( f ) can be applied with the roles ◦ ◦ of e and f interchanged; thus we obtain Bα ( f ) ⊆ Bα (e). Combining this with ◦ ◦ ◦ ◦ Bα (e) ⊆ Bα ( f ), we obtain Bα (e) = Bα ( f ). This proves Proposition 8.2. The Bhargava greedoid as a Gaussian elimination greedoid page 27

◦ ◦ ◦ Corollary 8.3. Let α ∈ V and e ∈ E. Let f ∈ Bα (e). Then, Bα (e) = Bα ( f ).

Proof. If e = f , then this is obvious. Hence, WLOG assume that e 6= f . Hence, ◦ d ( f , e) < α (since f ∈ Bα (e)). Thus, the “symmetry” axiom in Deﬁnition 2.3 yields ◦ ◦ that d (e, f ) = d ( f , e) < α. Hence, Proposition 8.2 yields Bα (e) = Bα ( f ). Qed.

Proposition 8.4. Let α ∈ V and e, f ∈ E be such that e 6= f and d (e, f ) > α. Then: ◦ ◦ (a) If a ∈ Bα (e) and b ∈ Bα ( f ), then a 6= b and d (a, b) > α. ◦ ◦ (b) The open balls Bα (e) and Bα ( f ) are disjoint. ◦ ◦ Proof. (a) Let a ∈ Bα (e) and b ∈ Bα ( f ). We must prove that a 6= b and d (a, b) > α. ◦ Assume the contrary. Thus, either a = b or else d (a, b) < α. Hence, a ∈ Bα (b) ◦ ◦ ◦ (by the definition of Bα (b)). Thus, Bα (b) = Bα (a) (by Corollary 8.3, applied to b ◦ ◦ ◦ and a instead of e and f ). Also, from a ∈ Bα (e), we obtain Bα (e) = Bα (a) (by ◦ Corollary 8.3, applied to a instead of f ). Furthermore, from b ∈ Bα ( f ), we obtain ◦ ◦ Bα ( f ) = Bα (b) (by Corollary 8.3, applied to f and b instead of e and f ). Finally, ◦ ◦ ◦ ◦ ◦ the definition of Bα (e) yields e ∈ Bα (e) = Bα (a) = Bα (b) = Bα ( f ). Since e 6= f , ◦ this entails d (e, f ) < α (by the definition of Bα ( f )). This contradicts d (e, f ) > α. This contradiction shows that our assumption was false. Hence, Proposition 8.4 (a) follows. (b) This is simply the “a 6= b” part of Proposition 8.4 (a). The next proposition shows how a finite V-ultra triple (of size > 1) can be de- composed into several smaller V-ultra triples; this will later be used for recursive reasoning:16

Proposition 8.5. Assume that E is ﬁnite and |E| > 1. Let α = max (d (E × E)). Pick any maximum-size α-clique, and write it in the form {e1, e2,..., em} for some distinct elements e1, e2,..., em of E. For each i ∈ {1, 2, . . . , m}, let E be the open ball B◦ (e ), and let (E , w , d ) be i α i i i i the V-ultra triple Ei, w |Ei , d |Ei×Ei . Then: (a) We have m > 1. (b) The sets E1, E2,..., Em form a set partition of E. (This means that these sets E1, E2,..., Em are disjoint and nonempty and their union is E.) (c) We have |Ei| < |E| for each i ∈ {1, 2, . . . , m}. (d) If i ∈ {1, 2, . . . , m}, and if a, b ∈ Ei are distinct, then d (a, b) < α. (e) If i and j are two distinct elements of {1, 2, . . . , m}, and if a ∈ Ei and b ∈ Ej, then a 6= b and d (a, b) = α. (f) Let ni = mcs (Ei, wi, di) for each i ∈ {1, 2, . . . , m}. Then,

mcs (E, w, d) = max {m, n1, n2,..., nm} .

16See Deﬁnition 4.4 for the meaning of mcs (E, w, d). The Bhargava greedoid as a Gaussian elimination greedoid page 28

Proof. Let us first check that α is well-defined. Indeed, the set E × E is nonempty (since |E| > 1) and finite (since E is finite). Hence, the set d (E × E) is nonempty and finite as well. Thus, its largest element max (d (E × E)) is well-defined. In other words, α is well-defined. We have d (a, b) 6 α for each (a, b) ∈ E × E (17) (since α = max (d (E × E))). The set {e1, e2,..., em} is a maximum-size α-clique (by its definition), but its size is m (since e1, e2,..., em are distinct). Hence, the maximum size of an α-clique is m. Thus, every α-clique has size 6 m. (a) From α = max (d (E × E)) ∈ d (E × E), we conclude that there exist two distinct elements u and v of E satisfying d (u, v) = α. Consider these u and v. Then, {u, v} is an α-clique. This α-clique must have size 6 m (since every α-clique has size 6 m). Hence, |{u, v}| 6 m. Thus, m > |{u, v}| = 2 (since u and v are distinct), so that m > 2 > 1. This proves Proposition 8.5 (a). (b) If i and j are two distinct elements of {1, 2, . . . , m}, then ei 6= ej (since e1, e2,..., em are distinct) and thus d ei, ej = α (since {e1, e2,..., em} is an α-clique), ◦ ◦ and therefore the open balls Bα (ei) and Bα ej are disjoint (by Proposition 8.4 (b), applied to e = ei and f = ej). In other words, if i and j are two distinct elements ◦ of {1, 2, . . . , m}, then the open balls Ei and Ej are disjoint (since Ei = Bα (ei) and ◦ Ej = Bα ej ). Hence, the open balls E1, E2,..., Em are disjoint. Furthermore, these balls are ◦ nonempty (since each open ball Ei = Bα (ei) contains at least the element ei). Furthermore, the union of these balls E1, E2,..., Em is the whole set E. [Proof: Assume the contrary. Thus, the union of the balls E1, E2,..., Em must be a proper subset of E (since it is clearly a subset of E). In other words, there exists some a ∈ E that belongs to none of these balls E1, E2,..., Em. Consider this a. Then, ◦ ◦ for each i ∈ {1, 2, . . . , m}, we have a ∈/ Ei = Bα (ei). By the definition of Bα (ei), this entails that a 6= ei and d (a, ei) > α, hence d (a, ei) = α (since (17) yields d (a, ei) 6 α). Thus, we have shown that d (a, ei) = α for all i ∈ {1, 2, . . . , m}. Therefore, {a, e1, e2,..., em} is an α-clique. This α-clique has size m + 1 (since e1, e2,..., em are distinct, and since a 6= ei for each i ∈ {1, 2, . . . , m}). But this contradicts the fact that every α-clique has size 6 m. This contradiction shows that our assumption was wrong. Hence, the union of the balls E1, E2,..., Em is the whole set E.] We have now proved that the balls E1, E2,..., Em are disjoint and nonempty and their union is the whole set E. In other words, they form a set partition of E. This proves Proposition 8.5 (b). (c) Proposition 8.5 (b) shows that the m sets E1, E2,..., Em form a set partition of E. Thus, these m sets are m disjoint nonempty subsets of E; hence, each of them is a proper subset of E (since m > 1). In other words, Ei is a proper subset of E for each i ∈ {1, 2, . . . , m}. Hence, |Ei| < |E| for each i ∈ {1, 2, . . . , m}. This proves Proposition 8.5 (c). (d) Let i ∈ {1, 2, . . . , m}. Let a, b ∈ Ei be distinct. We must prove that d (a, b) < α. The Bhargava greedoid as a Gaussian elimination greedoid page 29

◦ ◦ ◦ We have b ∈ Ei = Bα (ei) (by the definition of Ei) and thus Bα (ei) = Bα (b) (by ◦ ◦ Corollary 8.3, applied to ei and b instead of e and f ). But a ∈ Ei = Bα (ei) = Bα (b). ◦ In other words, a = b or else d (a, b) < α (by the definition of Bα (b)). Hence, d (a, b) < α (since a 6= b). This proves Proposition 8.5 (d). (e) Let i and j be two distinct elements of {1, 2, . . . , m}. Let a ∈ Ei and b ∈ Ej. We must prove that a 6= b and d (a, b) = α. ◦ ◦ We have a ∈ Ei = Bα (ei) (by the definition of Ei) and similarly b ∈ Bα ej . Furthermore, ei 6= ej (since i 6= j and since e1, e2,..., em are distinct) and thus d ei, ej = α (since {e1, e2,..., em} is an α-clique). Hence, Proposition 8.4 (a) (applied to ei and ej instead of e and f ) yields a 6= b and d (a, b) > α. Combining d (a, b) > α with (17), we obtain d (a, b) = α. This proves Proposition 8.5 (e). (f) Let us first notice that the map di (for each i ∈ {1, 2, . . . , m}) is defined to be a restriction of the map d. Thus, for any i ∈ {1, 2, . . . , m}, we have

di (e, f ) = d (e, f ) for any two distinct e, f ∈ Ei. (18)

The V-ultra triple (E, w, d) has a clique of size m (namely, {e1, e2,..., em}), and a clique of size ni for each i ∈ {1, 2, . . . , m} (indeed, ni = mcs (Ei, wi, di) shows that the V-ultra triple (Ei, wi, di) has such a clique; but that clique must of course be a clique of (E, w, d) as well17). Thus, the V-ultra triple (E, w, d) has a clique of size max {m, n1, n2,..., nm} (since max {m, n1, n2,..., nm} is one of the numbers m, n1, n2,..., nm). Thus,

mcs (E, w, d) > max {m, n1, n2,..., nm} .

It remains to prove the reverse inequality – i.e., to prove that mcs (E, w, d) 6 max {m, n1, n2,..., nm}. Assume the contrary. Thus, mcs (E, w, d) > max {m, n1, n2,..., nm}. The V-ultra triple (E, w, d) has a clique C of size mcs (E, w, d) (by the definition of mcs (E, w, d)). Consider this C. Thus, |C| = mcs (E, w, d) > max {m, n1, n2,..., nm} > 0. Hence, the set C is nonempty. In other words, there exists some a ∈ C. Consider this a. But recall that E1, E2,..., Em form a set partition of E. Thus, E1 ∪ E2 ∪ · · · ∪ Em = E. Now, a ∈ C ⊆ E = E1 ∪ E2 ∪ · · · ∪ Em. In other words, a ∈ Ei for some i ∈ {1, 2, . . . , m}. Consider this i. We are in one of the following two cases: Case 1: We have C ⊆ Ei. Case 2: We have C 6⊆ Ei. Let us first consider Case 1. In this case, we have C ⊆ Ei. Recall that C is a clique of mcs (E, w, d). Thus, from C ⊆ Ei, we conclude that C is a clique of the V-ultra triple (Ei, wi, di) (because of (18)). But the maximum size of such a clique is mcs (Ei, wi, di) (by the definition of mcs (Ei, wi, di)). Hence, |C| 6 mcs (Ei, wi, di) = ni. This contradicts |C| > max {m, n1, n2,..., nm} > ni. Thus, we have obtained a contradiction in Case 1. 17because of (18) The Bhargava greedoid as a Gaussian elimination greedoid page 30

Let us next consider Case 2. In this case, we have C 6⊆ Ei. In other words, there exists some b ∈ C such that b ∈/ Ei. Consider this b. From b ∈ C ⊆ E = E1 ∪ E2 ∪ · · · ∪ Em, we conclude that b ∈ Ej for some j ∈ {1, 2, . . . , m}. Consider this j. If we had i = j, then we would have b ∈ Ej = Ei (since j = i), which would contradict b ∈/ Ei. Hence, we cannot have i = j. Hence, i 6= j. Thus, Proposition 8.5 (e) yields that a 6= b and d (a, b) = α. Now, a 6= b shows that a and b are two distinct elements of C. But C is a clique, and thus is a γ-clique for some γ ∈ V. Consider this γ. Since C is a γ-clique, we have d (a, b) = γ (since a and b are two distinct elements of C). Hence, γ = d (a, b) = α. Thus, C is an α-clique (since C is a γ-clique). Thus, |C| 6 m (since every α-clique has size 6 m). This contradicts |C| > max {m, n1, n2,..., nm} > m. Thus, we have obtained a contradiction in Case 2. We have thus found a contradiction in each of the two Cases 1 and 2. Thus, we always have a contradiction. This completes the proof of Proposition 8.5 (f).

Remark 8.6. Here are some additional observations on Proposition 8.5, which we will not need (and thus will not prove): (a) The set partition {E1, E2,..., Em} constructed in Proposition 8.5 (b) does not depend on the choice of e1, e2,..., em. Indeed, E1, E2,..., Em are precisely the maximal (with respect to inclusion) subsets F of E satisfying (d (a, b) < α for any distinct a, b ∈ F). (b) Applying Proposition 8.5 iteratively, we can see that a V-ultra triple (E, w, d) with ﬁnite E has a recursive structure governed by a tree. This idea is not new; see [Lemin03] and [PetDov14, §2–§3] for related results.

9. Valadic representation of V-ultra triples

For the whole Section 9, we fix a field K, and we let V>0, L, L+ and tα be as in Section 6. We also recall Definition 6.7.

Deﬁnition 9.1. Let γ ∈ V and u ∈ L. We say that a valadic V-ultra triple (E, w, d) is (γ, u)-positioned if

E ⊆ u + t−γL+.

In other words, a valadic V-ultra triple (E, w, d) is (γ, u)-positioned if and only if each element of E has the form u + ∑ pβtβ for some pβ ∈ K. β∈V; β>−γ Theorem 9.2. Let (E, w, d) be a V-ultra triple such that the set E is ﬁnite. Let γ ∈ V be such that

d (a, b) 6 γ for each (a, b) ∈ E × E. (19) The Bhargava greedoid as a Gaussian elimination greedoid page 31

Let u ∈ L. Assume that |K| > mcs (E, w, d). Then, there exists a (γ, u)-positioned valadic V-ultra triple isomorphic to (E, w, d).

Proof of Theorem 9.2. Strong induction on |E|. If |E| = 1, then this is clear (just take the obvious valadic V-ultra triple on the set {u} ⊆ L, which is clearly (γ, u)-positioned). The case |E| = 0 is even more obvious. Thus, WLOG assume that |E| > 1. Thus, the set E × E is nonempty. Hence, d (E × E) is a nonempty ﬁnite subset of V, and therefore has a largest element. In other words, max (d (E × E)) is well-deﬁned. Let α = max (d (E × E)). Thus, α 6 γ (by (19)). Pick any maximum-size α-clique, and write it in the form {e1, e2,..., em} for some distinct elements e1, e2,..., em of E. For each i ∈ {1, 2, . . . , m}, let Ei be the open ◦ ball Bα (ei), and let (Ei, wi, di) be the V-ultra triple Ei, w |Ei , d |Ei×Ei . Then, Proposition 8.5 (a) shows that m > 1. Moreover, Proposition 8.5 (b) shows that the m sets E1, E2,..., Em form a set partition of E. Hence, E = E1 t E2 t · · · t Em ◦ (an internal disjoint union). Each set Ei is the open ball Bα (ei) and thus contains ei. Let ni = mcs (Ei, wi, di) for each i ∈ {1, 2, . . . , m}. Then,

|K| > mcs (E, w, d) = max {m, n1, n2,..., nm} (by Proposition 8.5 (f)) .

For any subset Z of L, we deﬁne a distance function dZ : Z × Z → V by setting

dZ (a, b) = − ord (a − b) for all (a, b) ∈ Z × Z. (20)

Note that the distance function of any valadic V-ultra triple is precisely dZ, where Z is the ground set of this V-ultra triple. We have |K| > max {m, n1, n2,..., nm} > m. Hence, there exist m distinct elements λ1, λ2,..., λm of K. Fix m such elements. Deﬁne m elements u1, u2,..., um of L by ui = u + λit−α for each i ∈ {1, 2, . . . , m} . (21)

Let L++ denote the K-submodule of L+ generated by tδ for all positive δ ∈ V. (Of course, δ ∈ V is said to be positive if and only if δ > 0.) It is easy to see that L++ is an ideal of L+. (Actually, L++ = Ker π, where π is as defined in Lemma 6.11.) Hence, L++L+ ⊆ L++. Let β be the second-largest element of d (E × E). (If this second-largest element does not exist, then we leave β undefined and should interpret t−β to mean 0 from now on.) The definition of β yields β < α (since α = max (d (E × E))). Hence, −β > −α, so that t−β ∈ t−αL++. (22)

(This holds even when β is undeﬁned, because t−β = 0 in this case.) Let i ∈ {1, 2, . . . , m}. We shall work under the assumption that β is well-deﬁned; we will later explain how to proceed without it. The Bhargava greedoid as a Gaussian elimination greedoid page 32

Let (a, b) ∈ Ei × Ei. Then, a and b are two distinct elements of Ei. Hence, Proposition 8.5 (d) yields

d (a, b) < α = max (d (E × E)) = (the largest element of d (E × E)) .

Hence, d (a, b) 6 (the second-largest element of d (E × E))(since d (a, b) ∈ d (E × E)) = β

(since the second-largest element of d (E × E) is β). Since di = d |Ei×Ei , we now have di (a, b) = d (a, b) 6 β. Forget that we ﬁxed (a, b). We thus have proved that

di (a, b) 6 β for each (a, b) ∈ Ei × Ei.

Also, |Ei| < |E| (by Proposition 8.5 (c)) and

|K| > max {m, n1, n2,..., nm} > ni = mcs (Ei, wi, di) .

Hence, the induction hypothesis shows that we can apply Theorem 9.2 to β, ui and (E , w , d ) instead of γ, u and (E, w, d). We thus conclude that there exists a (β, u )- i i i i positioned valadic V-ultra triple Ei, wi, di isomorphic to (Ei, wi, di). Consider this Ei, wi, di . The V-ultra triple (Ei, wi, di) is isomorphic to Ei, wi, di (since Ei, wi, di is isomorphic to (Ei, wi, di), but being isomorphic is a symmetric relation). In other words, there exists an isomorphism f : E → E of V-ultra triples i i i from (Ei, wi, di) to Ei, wi, di . Consider this fi. Since the V-ultra triple Ei, wi, di is (β, ui)-positioned, we have

Ei ⊆ ui + t−β L+ ⊆ ui + t−α L++L+ ⊆ ui + t−αL++. | {z } |{z} ⊆L ∈t−αL++ ++ (by (22))

Hence, we have found a V-ultra triple (E , w , d ) and a valadic V-ultra triple i i i Ei, wi, di satisfying

Ei ⊆ ui + t−αL++, and an isomorphism fi : Ei → Ei of V-ultra triples from (Ei, wi, di) to Ei, wi, di . We have done so assuming that β is well-deﬁned; but this is even easier when β is undeﬁned18.

18Proof. Assume that β is undeﬁned. In other words, the set d (E × E) has no second-largest element. Hence, all elements of d (E × E) are equal (since d (E × E) is a nonempty ﬁnite set). Thus, all elements of d (E × E) are α (since α = max (d (E × E))). In other words, d (a, b) = α The Bhargava greedoid as a Gaussian elimination greedoid page 33

Forget that we ﬁxed i. Thus, for each i ∈ {1, 2, . . . , m}, we have constructed a V-ultra triple (Ei, wi, di) and a valadic V-ultra triple Ei, wi, di satisfying

Ei ⊆ ui + t−αL++ and an isomorphism fi : Ei → Ei of V-ultra triples from (Ei, wi, di) to Ei, wi, di . For later use, let us observe the following:

Claim 1: Let i and j be two distinct elements of {1, 2, . . . , m}. Let a ∈ Ei and b ∈ Ej. Then, a − b 6= 0 and ord (a − b) = −α.

[Proof of Claim 1: From i 6= j, we conclude that λi and λj are two distinct elements of K (since λ1, λ2,..., λm are distinct elements of K). Hence, λi − λj is a nonzero element of K. We have a ∈ Ei ⊆ ui + t−αL++, so that a − ui ∈ t−αL++. Similarly, b − uj ∈ t−αL++. Also, (21) yields ui = u + λit−α. Likewise, uj = u + λjt−α. Subtracting the latter two equalities from one another, we obtain ui − uj = (u + λit−α) − u + λjt−α = λi − λj t−α.

Now, a − b = ui − uj + (a − ui) − b − uj | {z } | {z } | {z } ∈t−αL++ ∈t L =(λi−λj)t−α −α ++ ∈ λi − λj t−α + t−αL++ − t−αL++ ⊆ λi − λj t−α + t−αL++. | {z } ⊆t−αL++

In other words, a − b can be written in the form a − b = λi − λj t−α + a K-linear combination of tη with η > −α

for each (a, b) ∈ E × E. In other words, d (a, b) = α for any two distinct elements a and b of E. Thus, in particular, d (a, b) = α for any two distinct elements a and b of Ei. But Proposition 8.5 (d) shows that d (a, b) < α for any two distinct elements a and b of Ei. The previous two sentences would contradict each other if the set Ei had two distinct elements. Thus, the set Ei has no two distinct elements; in other words, |Ei| 6 1. Since ei is an element of Ei (because ◦ Ei = Bα (ei)), we thus have Ei = {ei}. Now, set Ei = {ui}; let wi : Ei → V be the map that sends ui to wi (ei); let di : Ei × Ei → V be the distance function d{u }; and let fi : Ei → Ei be i the map that sends ei to ui. Then, it is easy to see that Ei, wi, di is a valadic V-ultra triple

satisfying Ei ⊆ ui + t−αL++ (since Ei = {ui} = ui + 0 ⊆ ui + t−αL++), and that the map |{z} ⊆t−αL++ fi : Ei → Ei is an isomorphism of V-ultra triples from (Ei, wi, di) to Ei, wi, di (since Ei = {ei}

and Ei = {ui}, so that the maps di and di both have no values, whereas the map wi is deﬁned in such a way that wi (ui) = wi (ei) and thus wi ◦ fi = wi). Thus, everything we constructed above still exists when β is undeﬁned. The Bhargava greedoid as a Gaussian elimination greedoid page 34

(since the elements of t−αL++ are precisely the K-linear combinations of tη with η > −α). From this, we obtain two things: First, we obtain that [t−α](a − b) = λi − λj (since λi − λj ∈ K), hence [t−α](a − b) = λi − λj 6= 0 (since λi − λj is nonzero), and thus a − b 6= 0. Furthermore, we obtain that ord (a − b) = −α (again since λi − λj 6= 0). Thus, Claim 1 is proved.] Let E denote the subset E1 ∪ E2 ∪ · · · ∪ Em 19 of L. Note that the sets E1, E2,..., Em are disjoint . Hence, E = E1 t E2 t · · · t Em (an internal disjoint union). Moreover, each i ∈ {1, 2, . . . , m} satisﬁes

Ei ⊆ ui +t−αL++ = u + λit−α + t−αL++ |{z} | {z } =u+λit−α =t−α(λi+L++) (by (21))

= u + t−α (λi + L++) ⊆ u + t−γ L+L+ |{z} | {z } | {z } ∈t−γL+ ⊆L+ ⊆L+ (since −α>−γ (since λi∈K⊆L+) (because α6γ)) ⊆ u + t−γL+. (23) Now, m m [ [ E = E1 ∪ E2 ∪ · · · ∪ Em = Ei ⊆ (u + t−γL+) ⊆ u + t−γL+. i=1 |{z} i=1 ⊆u+t−γL+ (by (23))

Now, recall that E = E1 t E2 t · · · t Em and E = E1 t E2 t · · · t Em. Thus, we can glue the bijections fi : Ei → Ei together to a single bijection f : E → E,

a 7→ fi (a) , where i ∈ {1, 2, . . . , m} is such that a ∈ Ei. Consider this f . Deﬁne a weight function w : E → V by setting w = w ◦ f −1. Then, E, w, dE is a (γ, u)-positioned valadic V-ultra triple. (It is (γ, u)-positioned, since E ⊆ u + t−γL+.) Moreover, it is easy to see that every i, j ∈ {1, 2, . . . , m} and any two distinct elements a ∈ Ei and b ∈ Ej satisfy ( α, if i 6= j; d (a, b) = (24) E d (a, b) , if i = j Ei

19 Proof. Assume the contrary. Thus, there exist some distinct i, j ∈ {1, 2, . . . , m} such that Ei ∩ Ej 6= ∅. Consider these i and j. There exists some a ∈ Ei ∩ Ej (since Ei ∩ Ej 6= ∅). Consider this a. We have a ∈ Ei ∩ Ej ⊆ Ei and similarly a ∈ Ej. Hence, Claim 1 (applied to b = a) yields a − a 6= 0 and ord (a − a) = −α. But a − a 6= 0 clearly contradicts a − a = 0. This contradiction proves that our assumption was false, qed. The Bhargava greedoid as a Gaussian elimination greedoid page 35

20. From this, it is easy to see that

dE ( f (a) , f (b)) = d (a, b) for all (a, b) ∈ E × E (25) 21. Moreover, w ◦ f = w (since w = w ◦ f −1). Hence, the bijection f : E → E is an isomorphism of V-ultra triples from (E, w, d) to E, w, dE . Hence, there exists a (γ, u)-positioned valadic V-ultra triple isomorphic to (E, w, d) (namely, E, w, dE ). This proves Theorem 9.2 for our (E, w, d). Thus, the induction step is complete, and Theorem 9.2 is proven. 20 Proof of (24): Let i, j ∈ {1, 2, . . . , m}, and let a ∈ Ei and b ∈ Ej be two distinct elements. We must prove (24). It clearly sufﬁces to verify the equality ( −α, if i 6= j; ord (a − b) = ord (a − b) , if i = j

(since both d and d are given by the formula (20)). This equality is obviously true in the case E Ei when i = j; on the other hand, it follows from Claim 1 in the case when i 6= j. Hence, (24) is proved. 21Proof of (25): Let (a, b) ∈ E × E. Thus, a and b are two distinct elements of E. Let i, j ∈ {1, 2, . . . , m} be such that a ∈ Ei and b ∈ Ej. (These i and j clearly exist, because a and b both belong to E = E1 t E2 t · · · t Em.) Then, the deﬁnition of f yields f (a) = fi (a) and f (b) = fj (b). Hence, dE ( f (a) , f (b)) = dE fi (a) , fj (b) ( α, if i 6= j; = (26) d f (a) , f (b) , if i = j Ei i j

(by (24), applied to fi (a) and fj (b) instead of a and b). If i 6= j, then this becomes

dE ( f (a) , f (b)) = α = d (a, b)(since Proposition 8.5 (e) yields d (a, b) = α) , and thus (25) is proven in this case. Hence, for the rest of this proof of (25), we WLOG assume that i = j. Hence, b ∈ Ej = Ei (since j = i). Recall that fi : Ei → Ei is an isomorphism of V-ultra triples from (Ei, wi, di) to Ei, wi, di . Hence,

di ( fi (a) , fi (b)) = di (a, b) = d (a, b)(by the deﬁnition of di) . Also, d = d (since the V-ultra triple E , w , d is valadic). Now, (26) becomes i Ei i i i ( α, if i 6= j; d ( f (a) , f (b)) = E d f (a) , f (b) , if i = j Ei i j = d f (a) , f (b) (since i = j) Ei i j = d ( f (a) , f (b))(since j = i) Ei i i |{z} =di

= di ( fi (a) , fi (b)) = d (a, b) .

Thus, (25) is proven. The Bhargava greedoid as a Gaussian elimination greedoid page 36

10. Proof of the main theorem

We can now prove Theorem 4.7 (from which we will immediately obtain Theorem 3.1). Proof of Theorem 4.7. Pick some γ ∈ V such that

d (a, b) 6 γ for each (a, b) ∈ E × E.

(Such a γ clearly exists, since the set E × E is finite and the set V is totally ordered.) Theorem 9.2 (applied to u = 0) thus yields that there exists a (γ, 0)-positioned valadic V-ultra triple isomorphic to (E, w, d). Consider this valadic V-ultra triple, and denote it by (F, v, c). Let E denote the Bhargava greedoid of this V-ultra triple (F, v, c). The set F is a subset of L (since (F, v, c) is a valadic V-ultra triple) and is finite (since (F, v, c) is isomorphic to (E, w, d), whence |F| = |E| < ∞). Moreover, the distance function c of the V-ultra triple (F, v, c) is the function d from Definition 6.7 (since (F, v, c) is a valadic V-ultra triple). Hence, Theorem 6.9 (applied to (F, v, c) instead of (E, w, d)) yields that the Bhargava greedoid of (F, v, c) is the Gaussian elimination greedoid of a vector family over K. In other words, E is the Gaussian elimination greedoid of a vector family over K (since the Bhargava greedoid of (F, v, c) is E). But Proposition 7.4 yields that the Bhargava greedoids of (E, w, d) and (F, v, c) are isomorphic. In other words, the set systems F and E are isomorphic (since F and E are the Bhargava greedoids of (E, w, d) and (F, v, c), respectively). In other words, the set systems E and F are isomorphic. Hence, Proposition 7.5 shows that F is the Gaussian elimination greedoid of a vector family over K. This proves Theorem 4.7.

Proof of Theorem 3.1. We have |K| > |E| > mcs (E, w, d). Thus, Theorem 4.7 yields that F is the Gaussian elimination greedoid of a vector family over K. This proves Theorem 3.1.

11. Proof of Theorem 5.1

For the rest of Section 11, we ﬁx a V-ultra triple (E, w, d). Our next goal is to prove Theorem 5.1. We have to build several tools to this purpose.

11.1. Closed balls We will use a counterpart to the concept of open balls: the notion of closed balls. To wit, it is deﬁned as follows: The Bhargava greedoid as a Gaussian elimination greedoid page 37

Deﬁnition 11.1. Let α ∈ V and e ∈ E. The closed ball Bα (e) is deﬁned to be the subset { f ∈ E | f = e or else d ( f , e) 6 α} of E.

Clearly, for each α ∈ V and each e ∈ E, we have e ∈ Bα (e), so that the closed ball Bα (e) contains at least the element e. Most properties of open balls have analogues for closed balls. In particular, here is an analogue of Proposition 8.2:

Proposition 11.2. Let α ∈ V and e, f ∈ E be such that e 6= f and d (e, f ) 6 α. Then, Bα (e) = Bα ( f ).

Proof. This can be proved by a straightforward modiﬁcation of the above proof of Proposition 8.2 (namely, all “<” signs are replaced by “6” signs). Next comes an analogue of Corollary 8.3:

Corollary 11.3. Let α ∈ V and e ∈ E. Let f ∈ Bα (e). Then, Bα (e) = Bα ( f ).

Proof. This can be proved by a straightforward modiﬁcation of the above proof of Corollary 8.3 (namely, all “<” signs are replaced by “6” signs). Knowing these properties, we can easily obtain the following lemma:

Lemma 11.4. Let β ∈ V. Let C be a β-clique. (a) The closed balls Bβ (c) for all c ∈ C are identical. Now, let B = Bβ (c) for some c ∈ C. Then: (b) We have C ⊆ B. (c) For any distinct elements p, q ∈ B, we have d (p, q) 6 β. (d) For any n ∈ E \ B and any p, q ∈ B, we have d (n, p) = d (n, q).

(Intuitively, it helps to think of a clique C as an ultrametric analogue of a sphere, and of the set B constructed in Lemma 11.4 as being the whole closed ball whose boundary is this sphere. Of course, this must not be taken literally; in particular, every point in this disk serves as the “center” of this ball, so to speak.) Proof of Lemma 11.4. We know that C is a β-clique. In other words, C is a subset of E such that

any two distinct elements a, b ∈ C satisfy d (a, b) = β (27)

(by the definition of a “β-clique”). (a) We must prove that Bβ (e) = Bβ ( f ) for any e, f ∈ C. So let e, f ∈ C. We must prove that Bβ (e) = Bβ ( f ). If e = f , then this is obvious. Hence, we WLOG assume that e 6= f . Thus, (27) (applied to a = e and The Bhargava greedoid as a Gaussian elimination greedoid page 38 b = f ) yields d (e, f ) = β. Thus, the “symmetry” axiom in Definition 2.3 yields that d ( f , e) = d (e, f ) = β 6 β. Hence, f = e or else d ( f , e) 6 β. In other words, f ∈ Bβ (e) (by the definition of Bβ (e)). Hence, Corollary 11.3 (applied to α = β) yields Bβ (e) = Bβ ( f ). Hence, we have proved Bβ (e) = Bβ ( f ); thus, our proof of Lemma 11.4 (a) is complete. In preparation for the proofs of parts (b), (c) and (d), let us observe the following: We have defined B to be Bβ (c) for some c ∈ C. Consider this c. Thus, B = Bβ (c). Lemma 11.4 (a) says that

Bβ (a) = Bβ (b) for each a, b ∈ C. (28)

Thus, for each a ∈ C, we have

Bβ (a) = Bβ (c)(by (28), applied to b = c) = B since B = Bβ (c) . (29)

(b) Let a ∈ C. Then, a ∈ Bβ (a) (by the definition of Bβ (a), since a = a). Hence, a ∈ Bβ (a) = B (by (29)). Forget that we fixed a. We thus have proved that a ∈ B for each a ∈ C. In other words, C ⊆ B. This proves Lemma 11.4 (b). (c) Let p, q ∈ B be distinct. We must prove that d (p, q) 6 β. We have p ∈ B = Bβ (c). Hence, Corollary 11.3 (applied to α = β, e = c and f = p) yields Bβ (c) = Bβ (p). Now, q ∈ B = Bβ (c) = Bβ (p). In other words, we have q = p or else d (q, p) 6 β (by the definition of Bβ (p)). Since q = p is impossible (because p and q are distinct), we thus obtain d (q, p) 6 β. Thus, the “symmetry” axiom in Definition 2.3 yields that d (p, q) = d (q, p) 6 β. This proves Lemma 11.4 (c). (d) Let n ∈ E \ B and p, q ∈ B. We must prove that d (n, p) = d (n, q). If p = q, then this is obvious. Thus, we WLOG assume that p 6= q. Hence, Lemma 11.4 (c) yields d (p, q) 6 β. We have n ∈ E \ B. In other words, n ∈ E and n ∈/ B. But we have q ∈ B = Bβ (c). Hence, Corollary 11.3 (applied to α = β, e = c and f = q) yields Bβ (c) = Bβ (q). Now, n ∈/ B = Bβ (c) = Bβ (q). In other words, we don’t have (n = q or else d (n, q) 6 β) (by the definition of Bβ (q)). In other words, we have n 6= q and d (n, q) > β. Thus, d (n, q) > β > d (p, q) (since d (p, q) 6 β). We have p 6= n (since p ∈ B but n ∈/ B) and similarly q 6= n. Hence, the elements n, p and q of E are distinct (since p 6= n and q 6= n and p 6= q). The ultrametric triangle inequality (applied to n, p and q instead of a, b and c) thus yields

d (n, p) 6 max {d (n, q) , d (p, q)} = d (n, q)(since d (n, q) > d (p, q)) .

The same argument (with the roles of p and q interchanged) yields d (n, q) 6 d (n, p). Combining these two inequalities, we obtain d (n, p) = d (n, q). This proves Lemma 11.4 (d). The Bhargava greedoid as a Gaussian elimination greedoid page 39

11.2. Exchange results for sets intersecting a ball From now on, for the rest of Section 11, we assume that E is ﬁnite.

Corollary 11.5. Let β ∈ V. Let C be a β-clique. Let B = Bβ (c) for some c ∈ C. Let N be a subset of E \ B. Let P and Q be two subsets of B such that |P| = |Q|. Then: (a) We have PER (N ∪ Q) − PER (N ∪ P) = PER (Q) − PER (P). (b) Assume that the map w : E → V is constant. Assume further that Q is a subset of C. Then, PER (N ∪ Q) > PER (N ∪ P).

Proof of Corollary 11.5. Let m = |P| = |Q|. Let p1, p2,..., pm be all the m elements of P (listed without repetition). Let q1, q2,..., qm be all the m elements of Q (listed without repetition). (a) We have

d (n, pi) = d (n, qi) for each n ∈ N and i ∈ {1, 2, . . . , m} . (30)

[Proof of (30): Let n ∈ N and i ∈ {1, 2, . . . , m}. Then, n ∈ N ⊆ E \ B and pi ∈ P ⊆ B and qi ∈ Q ⊆ B. Thus, Lemma 11.4 (d) (applied to p = pi and q = qi) yields d (n, pi) = d (n, qi). This proves (30).] The set N is disjoint from B (since N is a subset of E \ B), and thus disjoint from P as well (since P ⊆ B). Hence, the deﬁnition of perimeter yields PER (N ∪ P) = PER (N) + PER (P) + ∑ ∑ d (n, p) n∈N p∈P | {z } m = ∑ d(n,pi) i=1 (since p1,p2,...,pm are all the m elements of P (listed without repetition)) m = PER (N) + PER (P) + ∑ ∑ d (n, pi) . n∈N i=1 Likewise, m PER (N ∪ Q) = PER (N) + PER (Q) + ∑ ∑ d (n, qi) . n∈N i=1 The Bhargava greedoid as a Gaussian elimination greedoid page 40

Subtracting the ﬁrst of these two equalities from the second, we obtain

PER (N ∪ Q) − PER (N ∪ P) m ! = PER (N) + PER (Q) + ∑ ∑ d (n, qi) n∈N i=1 m ! − PER (N) + PER (P) + ∑ ∑ d (n, pi) n∈N i=1 m m = PER (Q) − PER (P) + ∑ ∑ d (n, qi) − ∑ ∑ d (n, pi) n∈N i=1 n∈N i=1 | {z } =d(n,qi) (by (30)) m m = PER (Q) − PER (P) + ∑ ∑ d (n, qi) − ∑ ∑ d (n, qi) n∈N i=1 n∈N i=1 = PER (Q) − PER (P) .

This proves Corollary 11.5 (a). (b) We know that the map w : E → V is constant. Hence,

w (a) = w (b) for any a, b ∈ E. (31)

Each i ∈ {1, 2, . . . , m} satisﬁes

w (pi) = w (c) (32)

(by (31), applied to a = pi and b = c) and

w (qi) = w (c) (33)

(by (31), applied to a = qi and b = c). We know that C is a β-clique. In other words, C is a subset of E such that

any two distinct elements a, b ∈ C satisfy d (a, b) = β (34)

(by the deﬁnition of a “β-clique”). But p1, p2,..., pm are m distinct elements of P (by their deﬁnition). Hence, p1, p2,..., pm are m distinct elements of B (since P ⊆ B). Thus, if i and j are two distinct elements of {1, 2, . . . , m}, then pi and pj are two distinct elements of B, and therefore satisfy d pi, pj 6 β (35)

(by Lemma 11.4 (c), applied to p = pi and q = pj). On the other hand, q1, q2,..., qm are m distinct elements of Q (by their deﬁnition). Hence, q1, q2,..., qm are m distinct elements of C (since Q ⊆ C). Thus, if i and j are two distinct elements of {1, 2, . . . , m}, then qi and qj are two distinct elements of C, and therefore satisfy d qi, qj = β (36) The Bhargava greedoid as a Gaussian elimination greedoid page 41

(by (34), applied to a = qi and b = qj). Recall that p1, p2,..., pm are all the m elements of P (listed without repetition). Hence, the deﬁnition of perimeter yields m m PER (P) = ∑ w (pi) + ∑ d pi, pj 6 ∑ w (c) + ∑ β i=1 | {z } 1 i

(by the deﬁnition of Q, since q1, q2,..., qm are all the m elements of Q (listed without repetition)). But Corollary 11.5 (a) yields

PER (N ∪ Q) − PER (N ∪ P) = PER (Q) − PER (P) > 0 (since PER (P) 6 PER (Q)). In other words, PER (N ∪ Q) > PER (N ∪ P). This proves Corollary 11.5 (b).

Corollary 11.6. Let F be the Bhargava greedoid of (E, w, d). Assume that the map w : E → V is constant. Let β ∈ V. Let C be a β-clique. Let B = Bβ (c) for some c ∈ C. Let N be a subset of E \ B. Let P and Q be two subsets of B such that |P| = |Q| and Q ⊆ C and N ∪ P ∈ F. Then, N ∪ Q ∈ F.

Proof of Corollary 11.6. Corollary 11.5 (b) yields PER (N ∪ Q) > PER (N ∪ P). Also, the set N is disjoint from B (since N ⊆ E \ B), and thus is disjoint from P as well (since P is a subset of B). Hence, |N ∪ P| = |N| + |P|. The same argument (applied to Q instead of P) shows that |N ∪ Q| = |N| + |Q|. Hence, |N ∪ P| = |N| + |P| = |N| + |Q| = |N ∪ Q|. |{z} =|Q| We know that F is the Bhargava greedoid of (E, w, d). In other words, F = {A ⊆ E | A has maximum perimeter among all |A| -subsets of E} (37) (by Deﬁnition 2.7). Hence, from N ∪ P ∈ F, we conclude that the set N ∪ P has maximum perimeter among all |N ∪ P|-subsets of E. In other words, the set N ∪ P has maximum perimeter among all |N ∪ Q|-subsets of E (since |N ∪ P| = |N ∪ Q|). Since N ∪ Q is a further |N ∪ Q|-subset of E, we thus conclude that PER (N ∪ P) > PER (N ∪ Q). Combining this with PER (N ∪ Q) > PER (N ∪ P), we obtain PER (N ∪ Q) = PER (N ∪ P). In other words, the subsets N ∪ Q and N ∪ P of E have the same perimeter. Therefore, the set N ∪ Q has maximum perimeter among all |N ∪ Q|-subsets of E (because the set N ∪ P has maximum perimeter among all |N ∪ Q|-subsets of E). In view of (37), this entails that N ∪ Q ∈ F. This proves Corollary 11.6. The Bhargava greedoid as a Gaussian elimination greedoid page 42

11.3. Gaussian elimination greedoids in terms of determinants Next, we introduce some notations for matrices. Deﬁnition 11.7. Let n ∈ N and m ∈ N. Let A = ai,j be an n × m- 16i6n, 16j6m matrix. Let i1, i2,..., iu be some elements of {1, 2, . . . , n}; let j1, j2,..., jv be some j1,j2,...,jv elements of {1, 2, . . . , m}. Then, we deﬁne subi ,i ,...,i A to be the u × v-matrix 1 2 u aix,jy . 16x6u, 16y6v j1,j2,...,jv When i < i < ··· < iu and j < j < ··· < jv, the matrix sub A can 1 2 1 2 i1,i2,...,iu be obtained from A by crossing out all rows other than the i1-th, the i2-th, etc., the iu-th row and crossing out all columns other than the j1-th, the j2-th, etc., the j1,j2,...,jv jv-th column. Thus, in this case, sub A is called a submatrix of A. i1,i2,...,iu

 a b c d  1,3,4 Example 11.8. If n = 3 and m = 4 and A =  e f g h , then sub1,3 A = i j k ` a c d g f e (this is a submatrix of A) and sub3,2,1 A = (this is not, i k ` 2,3 k j i in general, a submatrix of A).

We can now describe Gaussian elimination greedoids in terms of determinants:

Lemma 11.9. Let n ∈ N. Let E be the set {1, 2, . . . , n}. m Let m ∈ N be such that m > |E|. Let K be a ﬁeld. For each e ∈ E, let ve ∈ K be a column vector. Let A be the m × n-matrix whose columns (from left to right) are v1, v2,..., vn. Let G be the Gaussian elimination greedoid of the vector family (ve) . e∈E Let p ∈ N. Let i1, i2,..., ip ∈ E be p distinct numbers. Let I = i1, i2,..., ip . Then, i1,i2,...,ip I ∈ G holds if and only if det sub1,2,...,p A 6= 0.

Proof of Lemma 11.9. We have I = i1, i2,..., ip . Thus, |I| = p (since i1, i2,..., ip are distinct) and I ⊆ E (since i1, i2,..., ip ∈ E). Define the maps πk for all k ∈ {0, 1, . . . , m} as in Definition 1.4. Since I ⊆ E, we have |I| 6 |E|. Hence, p = |I| 6 |E| 6 m (since m > |E|), so that p ∈ {0, 1, . . . , m}. m p Thus, the map πp : K → K is well-defined. Write the m × n-matrix A in the form A = ai,j . Then, the definition 16i6m, 16j6n i1,i2,...,ip of sub1,2,...,p A yields i1,i2,...,ip = sub1,2,...,p A ax,iy . 16x6p, 16y6p The Bhargava greedoid as a Gaussian elimination greedoid page 43

Thus, for each k ∈ {1, 2, . . . , p}, we have   a1,ik  a2,i  i1,i2,...,ip =  k  the k-th column of sub1,2,...,p A  .  . (38)  . 

ap,ik

The columns of the matrix A (from left to right) are v1, v2,..., vn. In other words, the `-th column of A is v` for each ` ∈ {1, 2, . . . , n}. In other words, for each ` ∈ {1, 2, . . . , n}, we have (the `-th column of A) = v`. Thus, for each ` ∈ {1, 2, . . . , n}, we have   a1,`  a `   2,  v` = (the `-th column of A) =  .   .  am,` (since A = ai,j ) and therefore 16i6m, 16j6n     a1,` a1,`  a `   a2,`   2,    πp (v`) = πp  .  =  .  (39)  .   .  am,` ap,`

(by the deﬁnition of πp). Hence, for each k ∈ {1, 2, . . . , p}, we have   a1,ik

 a2,ik  π v =   (by (39), applied to ` = i ) p ik  .  k  . 

ap,ik

i1,i2,...,ip = the k-th column of sub1,2,...,p A (by (38)) .

In other words, the vectors πp vi1 , πp vi2 ,..., πp vip are the columns of the i1,i2,...,ip matrix sub1,2,...,p A. We know that G is the Gaussian elimination greedoid of the vector family (ve)e∈E. Thus, Deﬁnition 1.4 shows that

F |F| G = F ⊆ E | the family π|F| (ve) ∈ K is linearly independent . e∈F The Bhargava greedoid as a Gaussian elimination greedoid page 44

Hence, we have the following chain of logical equivalences:

(I ∈ G) I |I| ⇐⇒ the family π|I| (ve) ∈ K is linearly independent e∈I ⇐⇒ ( ) ∈ (Kp)I the family πp ve e∈I is linearly independent (since |I| = p) ⇐⇒ the vectors πp (ve) for e ∈ I are linearly independent ⇐⇒ the vectors πp vi1 , πp vi2 ,..., πp vip are linearly independent   since the vectors πp (ve) for e ∈ I are precisely   the vectors πp vi1 , πp vi2 ,..., πp vip

i1,i2,...,ip ⇐⇒ the columns of the matrix sub1,2,...,p A are linearly independent   since the vectors πp vi , πp vi ,..., πp vi are  1 2 p   i1,i2,...,ip  the columns of the matrix sub1,2,...,p A

i1,i2,...,ip ⇐⇒ the matrix sub1,2,...,p A is invertible

i1,i2,...,ip ! since sub1,2,...,p A is a square matrix, and thus is invertible if and only if its columns are linearly independent

i1,i2,...,ip ⇐⇒ det sub1,2,...,p A 6= 0 .

This proves Lemma 11.9. We can leverage Lemma 11.9 to obtain a criterion that, roughly speaking, says that if a Gaussian elimination greedoid over a ﬁeld K contains a certain “constella- tion” (in an appropriate sense), then |K| must be > to a certain value. Namely:

Lemma 11.10. Let K be a ﬁeld. Let E be a ﬁnite set. Let F be the Gaussian elimination greedoid of a vector family (ve)e∈E over K. Let N and C be two disjoint subsets of E. Assume that the following three conditions hold:

(i) For any i ∈ C, we have N ∪ {i} ∈ F.

(ii) For any distinct i, j ∈ C, we have N ∪ {i, j} ∈ F.

(iii) For any p ∈ N and any distinct i, j ∈ C, we have (N ∪ {i, j}) \ {p} ∈/ F.

Then, |K| > |C|. The Bhargava greedoid as a Gaussian elimination greedoid page 45

m Proof of Lemma 11.10. Let m ∈ N be such that the vectors ve belong to K . Then, m > |E| (since otherwise, the Gaussian elimination greedoid of the vector family (ve)e∈E would not be well-deﬁned). Let n = |E| and r = |N|. Hence, N is an r-element subset of the n-element set E. Thus, the set E consists of the r elements of N and of the n − r remaining elements of E. Clearly, the claim we are proving will not change if we rename the elements of E. Thus, we can rename the elements of E arbitrarily. In particular, we can rename them in such a way that the r elements of the subset N will become 1, 2, . . . , r whereas all remaining n − r elements of E will become r + 1, r + 2, . . . , n. Thus, we can WLOG assume that the r elements of N are 1, 2, . . . , r and the remaining n − r elements of E are r + 1, r + 2, . . . , n. Assume this. Hence, N = {1, 2, . . . , r} and E \ N = {r + 1, r + 2, . . . , n} and therefore E = {1, 2, . . . , n}. Since the subsets N and C of E are disjoint, we have C ⊆ E \ N = {r + 1, r + 2, . . . , n}. Let A be the m × n-matrix whose columns (from left to right) are v , v ,..., vn. 1 2 Write this matrix A in the form A = ai,j . 16i6m, 16j6n We now make a few observations:

1,2,...,r,i Claim 1: We have det sub1,2,...,r+1 A 6= 0 for each i ∈ C.

[Proof of Claim 1: Let i ∈ C. Then, i ∈ C ⊆ {r + 1, r + 2, . . . , n}. Thus, the r + 1 numbers 1, 2, . . . , r, i are distinct. Also, N ∪ {i} ∈ F (by condition (i) in Lemma 11.10). But N ∪ {i} = {1, 2, . . . , r, i} (since N = {1, 2, . . . , r}). Thus, Lemma 11.9 (applied to F, N ∪ {i}, r + 1 and (1, 2, . . . , r, i) instead of G, I, p and i1, i2,..., ip ) 1,2,...,r,i yields that N ∪ {i} ∈ F holds if and only if det sub1,2,...,r+1 A 6= 0. Thus, we have 1,2,...,r,i det sub1,2,...,r+1 A 6= 0 (since N ∪ {i} ∈ F). This proves Claim 1.] Now, for each i ∈ C, we deﬁne a scalar ri ∈ K by

ar+2,i ri = . (40) 1,2,...,r,i det sub1,2,...,r+1 A

1,2,...,r,i This is well-deﬁned, since Claim 1 yields det sub1,2,...,r+1 A 6= 0.

Claim 2: The scalars ri for all i ∈ C are distinct.

[Proof of Claim 2: We need to prove that ri 6= rj for any two distinct i, j ∈ C. So let us ﬁx two distinct i, j ∈ C. We must prove that ri 6= rj. We have i, j ∈ C ⊆ {r + 1, r + 2, . . . , n}. Thus, the r + 2 numbers 1, 2, . . . , r, i, j are distinct (since i and j are distinct). Also, N ∪ {i, j} ∈ F (by condition (ii) in Lemma 11.10). But N ∪ {i, j} = {1, 2, . . . , r, i, j} (since N = {1, 2, . . . , r}). Thus, Lemma 11.9 (applied to F, N ∪ {i, j}, r + 2 and (1, 2, . . . , r, i, j) instead of G, I, p and The Bhargava greedoid as a Gaussian elimination greedoid page 46

1,2,...,r,i,j i1, i2,..., ip ) yields that N ∪ {i, j} ∈ F holds if and only if det sub1,2,...,r+2 A 6= 0. Thus, we have 1,2,...,r,i,j det sub1,2,...,r+2 A 6= 0 (41) (since N ∪ {i, j} ∈ F). Let us agree to use the following notation: If p ∈ {1, 2, . . . , r} is arbitrary, then “1, 2, . . . , pb,..., r, i, j” will denote the list 1, 2, . . . , r, i, j with the entry p omitted (i.e., the list 1, 2, . . . , p − 1, p + 1, p + 2, . . . , r, i, j). Now, let us use Laplace expansion to expand the determinant of the (r + 2) × 1,2,...,r,i,j (r + 2)-matrix sub1,2,...,r+2 A along its last row. We thus obtain

1,2,...,r,i,j det sub1,2,...,r+2 A r = − (r+2)+p 1,2,...,pb,...,r,i,j ∑ ( 1) ar+2,p det sub1,2,...,r+1 A p=1 (r+2)+(r+1) 1,2,...,r,j + (−1) ar+2,i det sub1,2,...,r+1 A (r+2)+(r+2) 1,2,...,r,i + (−1) ar+2,j det sub1,2,...,r+1 A . (42)

1,2,...,r,i,j (Indeed, the entries of the last row of sub1,2,...,r+2 A are

ar+2,1, ar+2,2,..., ar+2,r , ar+2,i, ar+2,j, | {z } these are the ar+2,p for all p∈{1,2,...,r} and the cofactors corresponding to the ﬁrst r of these entries are (r+2)+p 1,2,...,pb,...,r,i,j (−1) det sub1,2,...,r+1 A for all p ∈ {1, 2, . . . , r} , whereas the cofactors corresponding to the last two entries are

(r+2)+(r+1) 1,2,...,r,j (r+2)+(r+2) 1,2,...,r,i (−1) det sub1,2,...,r+1 A and (−1) det sub1,2,...,r+1 A .

) r Now, let us simplify the entries in the ∑ sum on the right hand side of (40). p=1 Let p ∈ {1, 2, . . . , r}. Then, p ∈ {1, 2, . . . , r} = N. Hence, (N ∪ {i, j}) \ {p} ∈/ F (by condition (iii) in Lemma 11.10). In other words, we don’t have (N ∪ {i, j}) \ {p} ∈ F. But from N ∪ {i, j} = {1, 2, . . . , r, i, j}, we obtain (N ∪ {i, j}) \ {p} = {1, 2, . . . , r, i, j} \ {p} = {1, 2, . . . , pb,..., r, i, j} (since the r + 2 numbers 1, 2, . . . , r, i, j are distinct). Of course, the r + 1 numbers 1, 2, . . . , pb,..., r, i, j are distinct (since the r + 2 numbers 1, 2, . . . , r, i, j are distinct). Thus, Lemma 11.9 (applied to F, (N ∪ {i, j}) \ {p}, r + 1 and (1, 2, . . . , p,..., r, i, j) instead of G, I, p and i1, i2,..., ip ) b 1,2,...,pb,...,r,i,j yields that (N ∪ {i, j}) \ {p} ∈ F holds if and only if det sub1,2,...,r+1 A 6= 0. The Bhargava greedoid as a Gaussian elimination greedoid page 47

1,2,...,pb,...,r,i,j Thus, we don’t have det sub1,2,...,r+1 A 6= 0 (since we don’t have (N ∪ {i, j}) \ {p} ∈ F). In other words, we have 1,2,...,pb,...,r,i,j det sub1,2,...,r+1 A = 0. (43)

Forget that we ﬁxed p. We thus have proved (43) for each p ∈ {1, 2, . . . , r}. Now, (42) becomes

1,2,...,r,i,j det sub1,2,...,r+2 A r = − (r+2)+p 1,2,...,pb,...,r,i,j ∑ ( 1) ar+2,p det sub1,2,...,r+1 A p=1 | {z } =0 (by (43)) (r+2)+(r+1) 1,2,...,r,j + (−1) ar+2,i det sub1,2,...,r+1 A | {z } =−1 (r+2)+(r+2) 1,2,...,r,i + (−1) ar+2,j det sub1,2,...,r+1 A | {z } =1 r = − (r+2)+p − 1,2,...,r,j + 1,2,...,r,i ∑ ( 1) ar+2,p0 ar+2,i det sub1,2,...,r+1 A ar+2,j det sub1,2,...,r+1 A p=1 | {z } =0 1,2,...,r,j = − ar+2,i det sub1,2,...,r+1 A | {z } 1,2,...,r,i =det(sub1,2,...,r+1 A)·ri (by (40)) 1,2,...,r,i + ar+2,j det sub1,2,...,r+1 A | {z } 1,2,...,r,j =det sub1,2,...,r+1 A ·rj (since (40) (applied to j instead of i) ar+2,j yields r = ) j 1,2,...,r,j det sub1,2,...,r+1 A 1,2,...,r,i 1,2,...,r,j = − det sub1,2,...,r+1 A · ri · det sub1,2,...,r+1 A 1,2,...,r,j 1,2,...,r,i + det sub1,2,...,r+1 A · rj · det sub1,2,...,r+1 A 1,2,...,r,i 1,2,...,r,j = det sub1,2,...,r+1 A det sub1,2,...,r+1 A · rj − ri . Hence,

1,2,...,r,i 1,2,...,r,j 1,2,...,r,i,j det sub1,2,...,r+1 A det sub1,2,...,r+1 A · rj − ri = det sub1,2,...,r+2 A 6= 0

(by (41)). Thus, rj − ri 6= 0, so that ri 6= rj. This proves Claim 2.] The Bhargava greedoid as a Gaussian elimination greedoid page 48

Claim 2 shows that the scalars ri for all i ∈ C are distinct. Thus, we have found |C| distinct elements of K (namely, these scalars ri for all i ∈ C). Therefore, K must have at least |C| elements. In other words, |K| > |C|. Thus, Lemma 11.10 is proved.

11.4. Proving the theorem We shall need one more lemma about Gaussian elimination greedoids:

Lemma 11.11. Let K be a ﬁeld. Let F be the Gaussian elimination greedoid of a vector family over K. If B ∈ F satisﬁes |B| > 0, then there exists b ∈ B such that B \ {b} ∈ F.

Proof of Lemma 11.11. We shall use the notion of a “strong greedoid”, as defined in Definition 12.1. Theorem 1.6 shows that every Gaussian elimination greedoid is a strong greedoid. Hence, F is a strong greedoid. In other words, F is a set system that satisfies the four axioms (i), (ii), (iii) and (iv) of Definition 12.1. Thus, in particular, F satisfies axiom (ii). But this axiom says precisely that if B ∈ F satisfies |B| > 0, then there exists b ∈ B such that B \ {b} ∈ F. Thus, Lemma 11.11 is proved. We can now prove Theorem 5.1: Proof of Theorem 5.1. The definition of mcs (E, w, d) shows that mcs (E, w, d) is the maximum size of a clique of (E, w, d). Thus, there exists a clique C of size mcs (E, w, d). Consider this C. Thus, mcs (E, w, d) = |C| (since C has size mcs (E, w, d)). We must prove that |K| > mcs (E, w, d). In other words, we must prove that |K| > |C| (since mcs (E, w, d) = |C|). If |C| 6 1, then this is obvious (since |K| > 1). Thus, we WLOG assume that |C| > 1. Hence, |C| > 2. The set C is a clique. In other words, there exists a β ∈ V such that C is a β-clique. Consider this β. Choose some c ∈ C. (We can do this, since |C| > 2 > 0.) Set B = Bβ (c). Lemma 11.4 (b) yields C ⊆ B, so that B ⊇ C. A secant set shall mean a subset S of E satisfying S ∈ F and |S ∩ B| > 2. The set E itself satisfies E ∈ F (by Remark 2.10) and |E ∩ B| > 2 22. In other words, E is a secant set. Hence, there exists at least one secant set. Thus, there exists a secant set of smallest size (since there are only finitely many secant sets). Consider such a secant set, and call it S. Thus, S is a secant set of smallest size. Hence, S is a secant set; in other words, S is a subset of E satisfying S ∈ F and |S ∩ B| > 2. Since S ∩ B is a subset of S, we have |S ∩ B| 6 |S|, so that |S| > |S ∩ B| > 2 > 0. Thus, Lemma 11.11 (applied to B = S) yields that there exists b ∈ S such that S \ {b} ∈ F. Consider this b. We now shall show several claims:

22Proof. Since B is a subset of E, we have E ∩ B = B ⊇ C. Thus, |E ∩ B| > |C| > 2. The Bhargava greedoid as a Gaussian elimination greedoid page 49

Claim 1: We have |(S ∩ B) \ {b}| < 2. [Proof of Claim 1: The set S \ {b} has smaller size than S (since b ∈ S), and thus cannot be a secant set (since S is a secant set of smallest size). In other words, S \ {b} cannot be a subset of E satisfying S \ {b} ∈ F and |(S \ {b}) ∩ B| > 2 (by the deﬁnition of “secant set”). Hence, we cannot have |(S \ {b}) ∩ B| > 2 (because S \ {b} is a subset of E satisfying S \ {b} ∈ F). In other words, we have |(S \ {b}) ∩ B| < 2. But it is known from basic set theory that (X ∩ Y) \ Z = (X \ Z) ∩ Y for any three sets X, Y and Z. Applying this to X = S, Y = B and Z = {b}, we obtain (S ∩ B) \ {b} = (S \ {b}) ∩ B. Hence, |(S ∩ B) \ {b}| = |(S \ {b}) ∩ B| < 2. This proves Claim 1.] Claim 2: We have b ∈ S ∩ B. [Proof of Claim 2: Claim 1 yields |(S ∩ B) \ {b}| < 2 6 |S ∩ B| (since |S ∩ B| > 2). Hence, |(S ∩ B) \ {b}| 6= |S ∩ B|, so that (S ∩ B) \ {b} 6= S ∩ B. Therefore, b ∈ S ∩ B. This proves Claim 2.] Claim 3: We have |S ∩ B| = 2. [Proof of Claim 3: Claim 1 says that |(S ∩ B) \ {b}| < 2. But Claim 2 yields b ∈ S ∩ B. Thus, |(S ∩ B) \ {b}| = |S ∩ B| − 1. Hence, |S ∩ B| = |(S ∩ B) \ {b}| +1 < | {z } <2 2 + 1 = 3. In other words, |S ∩ B| 6 2 (since |S ∩ B| is an integer). Combining this with |S ∩ B| > 2, we ﬁnd |S ∩ B| = 2. This proves Claim 3.] Claim 3 shows that the set S ∩ B has exactly two elements. One of these two elements is b (since Claim 2 says that b ∈ S ∩ B); let a be the other element. Thus, a 6= b and S ∩ B = {a, b}. Hence, {a, b} = S ∩ B ⊆ B. Also, |{a, b}| = 2 (since a 6= b). Let N = S \ B. Then, it is easy to see that S = N ∪ {a, b} 23 and N ∩ {a, b} = ∅ 24. From S = N ∪ {a, b}, we obtain |S| = |N ∪ {a, b}| = |N| + |{a, b}| (since N ∩ {a, b} = ∅) | {z } =2 = |N| + 2. (44) Moreover, N = S \B ⊆ E \ B ⊆ E \ C. |{z} |{z} ⊆E ⊇C Hence, the two subsets N and C of E are disjoint. Now, we have the following:

23Proof. Any two sets X and Y satisfy X = (X \ Y) ∪ (X ∩ Y). Applying this to X = S and Y = B, we obtain S = (S \ B) ∪ (S ∩ B) = N ∪ {a, b}. | {z } | {z } =N ={a,b} 24Proof. We have {a, b} ⊆ B. Hence, N ∩ {a, b} ⊆ (S \ B) ∩ B = ∅, so that N ∩ {a, b} = ∅. |{z} | {z } =S\B ⊆B The Bhargava greedoid as a Gaussian elimination greedoid page 50

Claim 4: Let i ∈ C. Then, N ∩ {i} = ∅ and N ∪ {i} ∈ F. [Proof of Claim 4: If we had i ∈ N, then we would have i ∈/ B (since i ∈ N = S \ B), which would contradict i ∈ C ⊆ B. Hence, we cannot have i ∈ N. Thus, N ∩ {i} = ∅. It remains to prove that N ∪ {i} ∈ F. If we had b ∈ N, then we would have b ∈/ B (since b ∈ N = S \ B), which would contradict b ∈ {a, b} ⊆ B. Hence, we cannot have b ∈ N. Thus, b ∈/ N, so that N \ {b} = N. We have S = N ∪ {a, b}, thus S \ {b} = (N ∪ {a, b}) \ {b} = (N \ {b}) ∪ ({a, b} \ {b}) = N ∪ {a} . | {z } | {z } =N ={a} (since a6=b) Hence, N ∪ {a} = S \ {b} ∈ F. But {a} and {i} are subsets of B (since a ∈ {a, b} ⊆ B and i ∈ C ⊆ B), and satisfy |{a}| = |{i}| (since both |{a}| and |{i}| equal 1) and {i} ⊆ C (since i ∈ C) and N ∪ {a} ∈ F. Hence, Corollary 11.6 (applied to P = {a} and Q = {i}) yields N ∪ {i} ∈ F. This ﬁnishes the proof of Claim 4.]

Claim 5: Let i, j ∈ C be distinct. Then, N ∩ {i, j} = ∅ and N ∪ {i, j} ∈ F. [Proof of Claim 5: If we had i ∈ N, then we would have i ∈/ B (since i ∈ N = S \ B), which would contradict i ∈ C ⊆ B. Hence, we cannot have i ∈ N. In other words, i ∈/ N. Likewise, j ∈/ N. Combining i ∈/ N with j ∈/ N, we obtain N ∩ {i, j} = ∅. It remains to prove that N ∪ {i, j} ∈ F. Note that |{a, b}| = 2 and |{i, j}| = 2 (since i and j are distinct). Hence, |{a, b}| = 2 = |{i, j}|. Also, i, j ∈ C, so that {i, j} ⊆ C ⊆ B. We have S = N ∪ {a, b}, thus N ∪ {a, b} = S ∈ F. But {a, b} and {i, j} are subsets of B (since {a, b} ⊆ B and {i, j} ⊆ B), and satisfy |{a, b}| = |{i, j}| and {i, j} ⊆ C and N ∪ {a, b} ∈ F. Hence, Corollary 11.6 (applied to P = {a, b} and Q = {i, j}) yields N ∪ {i, j} ∈ F. This ﬁnishes the proof of Claim 5.]

Claim 6: Let p ∈ N. Let i, j ∈ C be distinct. Then, (N ∪ {i, j}) \ {p} ∈/ F.

[Proof of Claim 6: Assume the contrary. Hence, (N ∪ {i, j}) \ {p} ∈ F. Also, |((N ∪ {i, j}) \ {p}) ∩ B| > 2 25. Therefore, (N ∪ {i, j}) \ {p} is a secant set (by the deﬁnition of “secant set”). Hence,

|(N ∪ {i, j}) \ {p}| > |S| (45) 25Proof. We have p ∈ N = S \ B, so that p ∈/ B. Hence, B \ {p} = B. Also, i, j ∈ C, so that {i, j} ⊆ C ⊆ B. But it is known from set theory that (X \ Y) ∩ Z = X ∩ (Z \ Y) for any three sets X, Y and Z. Applying this to X = N ∪ {i, j}, Y = {p} and Z = B, we obtain

((N ∪ {i, j}) \ {p}) ∩ B = (N ∪ {i, j}) ∩ (B \ {p}) ⊇ {i, j} ∩ B = {i, j} | {z } | {z } ⊇{i,j} =B

(since {i, j} ⊆ B). Hence, |((N ∪ {i, j}) \ {p}) ∩ B| > |{i, j}| = 2 (since i and j are distinct). Qed. The Bhargava greedoid as a Gaussian elimination greedoid page 51

(since S was deﬁned to be a secant set of smallest size). We have |{i, j}| = 2 (since i and j are distinct). But Claim 5 yields N ∩ {i, j} = ∅, so that |N ∪ {i, j}| = |N| + |{i, j}| = |N| + 2 = |S| (by (44)). But p ∈ N ⊆ N ∪ {i, j} | {z } =2 and thus |(N ∪ {i, j}) \ {p}| = |N ∪ {i, j}| −1 = |S| − 1 < |S|. This contradicts (45). | {z } =|S| This contradiction shows that our assumption was false. Hence, Claim 6 is proved.] But recall that F is the Gaussian elimination greedoid of a vector family over K (by assumption). Let (ve)e∈E be this vector family. Recall that N and C are two disjoint subsets of E. Moreover, the following facts hold: (i) For any i ∈ C, we have N ∪ {i} ∈ F (by Claim 4). (ii) For any distinct i, j ∈ C, we have N ∪ {i, j} ∈ F (by Claim 5). (iii) For any p ∈ N and any distinct i, j ∈ C, we have (N ∪ {i, j}) \ {p} ∈/ F (by Claim 6).

Hence, Lemma 11.10 shows that |K| > |C|. This proves Theorem 5.1.

12. Appendix: Gaussian elimination greedoids are strong

We shall now prove Theorem 1.6. In order to do so, we will first need to recall the definition of a strong greedoid (see, e.g., [GriPet19, §6.1]): Definition 12.1. Let E be a finite set. A set system F on ground set E is said to be a greedoid if it satisfies the following three axioms:

(i) We have ∅ ∈ F. (ii) If B ∈ F satisﬁes |B| > 0, then there exists b ∈ B such that B \ {b} ∈ F.

(iii) If A, B ∈ F satisfy |B| = |A| + 1, then there exists b ∈ B \ A such that A ∪ {b} ∈ F.

Furthermore, F is said to be a strong greedoid if it satisﬁes the following axiom (in addition to axioms (i), (ii) and (iii)):

(iv) If A, B ∈ F satisfy |B| = |A| + 1, then there exists b ∈ B \ A such that A ∪ {b} ∈ F and B \ {b} ∈ F.

Note that axiom (iii) in Deﬁnition 12.1 clearly follows from axiom (iv); thus, only axioms (i), (ii) and (iv) need to be checked in order to convince ourselves that a set system is a strong greedoid. The Bhargava greedoid as a Gaussian elimination greedoid page 52

We shall furthermore use a determinantal identity due to Plücker (one of several facts known as the Plücker identities). To state this identity, we will use the following notations:

Deﬁnition 12.2. Let K be a commutative ring. Let n ∈ N and m ∈ N. Let A ∈ Kn×m be an n × m-matrix. (Here and in the following, Kn×m denotes the set of all n × m-matrices with entries in K.) (a) If v ∈ Kn×1 is a column vector with n entries, then (A | v) will denote the n × (m + 1)-matrix whose m + 1 columns are A•,1, A•,2,..., A•,m, v (from left to right). (Informally speaking, (A | v) is the matrix obtained when the column vector v is “attached” to A at the right edge.) (b) If j ∈ {1, 2, . . . , m}, then A•,j shall mean the j-th column of the matrix A. This is a column vector with n entries, i.e., an n × 1-matrix. (c) If i ∈ {1, 2, . . . , n}, then A∼i,• shall mean the matrix obtained from the matrix A by removing the i-th row. This is an (n − 1) × m-matrix. (d) If i ∈ {1, 2, . . . , n} and j ∈ {1, 2, . . . , m}, then A∼i,∼j shall mean the matrix obtained from A by removing the i-th row and the j-th column. This is an (n − 1) × (m − 1)-matrix.

We can now state our Plücker identity:

Proposition 12.3. Let K be a commutative ring. Let n be a positive integer. Let X ∈ Kn×(n−1) and Y ∈ Kn×n. Let i ∈ {1, 2, . . . , n}. Then,

n n+q det (X∼i,•) det Y = ∑ (−1) det X | Y•,q det Y∼i,∼q . q=1

 a a0   x x0 x00  Example 12.4. If we set n = 3, X =  b b0 , Y =  y y0 y00  and i = 2, c c0 z z0 z00 then Proposition 12.3 states that

 x x0 x00  a a0 det det  y y0 y00  c c0 z z0 z00  a a0 x   a a0 x0  x0 x00 x x00 = det  b b0 y  det − det  b b0 y0  det z0 z00 z z00 c c0 z c c0 z0  a a0 x00  x x0 + det  b b0 y00  det . z z0 c c0 z00 The Bhargava greedoid as a Gaussian elimination greedoid page 53

Proposition 12.3 is precisely [Grinbe15, Proposition 6.137] (with A, C and v renamed as X, Y and i). Proof of Theorem 1.6. We need to prove that G is a strong greedoid. In other words, we need to prove that the axioms (i), (ii), (iii) and (iv) from Definition 12.1 are satisfied for F = G. Let us do this: ∅ |∅| Proof of the axiom (i) for F = G: The vector family π|∅| (ve) ∈ K is e∈∅ empty, and thus is linearly independent. In other words, ∅ ∈ G. This proves that the axiom (i) from Definition 12.1 is satisfied for F = G. Proof of the axiom (ii) for F = G: Let B ∈ G satisfy |B| > 0. We shall show that there exists b ∈ B such that B \ {b} ∈ G. Let n = |B|. Thus, n = |B| > 0. B |B| We have B ∈ G. In other words, the family π|B| (ve) ∈ K is linearly e∈B independent (by the definition of G). In other words, the family (πn (ve))e∈B ∈ (Kn)B is linearly independent (since |B| = n). Let Y be the n × n-matrix whose columns are the vectors πn (ve) for all e ∈ B (in some order, with no repetition). The columns of this n × n-matrix Y are thus B |B| linearly independent (since the family (πn (ve))e∈B ∈ K is linearly independent). Hence, this n × n-matrix Y is invertible. In other words, det Y 6= 0. Now, let yi,j be the (i, j)-th entry of the matrix Y for each i, j ∈ {1, 2, . . . , n}. Then, Laplace expansion along the n-th row yields

n n+q det Y = ∑ (−1) yn,q det Y∼n,∼q . q=1 If every q ∈ {1, 2, . . . , n} satisﬁed det Y∼n,∼q = 0, then this would become

n n+q det Y = ∑ (−1) yn,q det Y∼n,∼q = 0, q=1 | {z } =0 which would contradict det Y 6= 0. Hence, not every q ∈ {1, 2, . . . , n} satisfies det Y∼n,∼q = 0. In other words, there exists at least one q ∈ {1, 2, . . . , n} such that det Y∼n,∼q 6= 0. Consider this q. The q-th column of Y is the vector πn v f for some f ∈ B (by the definition of Y); consider this f . Note that |B \ { f }| = |B| − 1 = n − 1 (since |B| = n). The (n − 1) × (n − 1)-matrix Y∼n,∼q is invertible (since det Y∼n,∼q 6= 0); thus, its n − 1 columns are linearly independent. But these n − 1 columns are precisely the n − 1 vectors πn−1 (ve) for all e ∈ B \ { f } (by the definition of Y and the choice of f ). Hence, the n − 1 vectors πn−1 (ve) for all e ∈ B \ { f } are linearly independent. In other words, the family (πn−1 (ve))e∈B\{ f } is linearly independent. In other words, the family π|B\{ f }| (ve) is linearly independent (since |B \ { f }| = n − 1). e∈B\{ f } The Bhargava greedoid as a Gaussian elimination greedoid page 54

In other words, B \ { f } ∈ G (by the definition of G). Hence, there exists b ∈ B such that B \ {b} ∈ G (namely, b = f ). This shows that the axiom (ii) from Definition 12.1 is satisfied for F = G. Proof of the axiom (iv) for F = G: Let A, B ∈ G satisfy |B| = |A| + 1. We shall show that there exists b ∈ B \ A such that A ∪ {b} ∈ G and B \ {b} ∈ G. Let n = |B|. Thus, n = |B| = |A| + 1 > 1 > 0. Also, from n = |A| + 1, we obtain |A| = n − 1. B |B| We have B ∈ G. In other words, the family π|B| (ve) ∈ K is linearly e∈B independent (by the definition of G). In other words, the family (πn (ve))e∈B ∈ n B (K ) is linearly independent (since |B| = n). Similarly, the family (πn−1 (ve))e∈A ∈ A Kn−1 is linearly independent. Let Y be the n × n-matrix whose columns are the vectors πn (ve) for all e ∈ B (in some order, with no repetition). The columns of this n × n-matrix Y are thus B |B| linearly independent (since the family (πn (ve))e∈B ∈ K is linearly independent). Hence, this n × n-matrix Y is invertible. In other words, det Y 6= 0. Let X be the n × (n − 1)-matrix whose columns are the vectors πn (ve) for all e ∈ A (in some order, with no repetition). Then, the columns of the (n − 1) × (n − 1)- matrix X∼n,• are the vectors πn−1 (ve) for all e ∈ A; thus, they are linearly indepen- n−1A dent (since the family (πn−1 (ve))e∈A ∈ K is linearly independent). Hence, this (n − 1) × (n − 1)-matrix X∼n,• is invertible. In other words, det (X∼n,•) 6= 0. But K is a field and thus an integral domain. Hence, from det (X∼n,•) 6= 0 and det Y 6= 0, we obtain det (X∼n,•) det Y 6= 0. (46) If every q ∈ {1, 2, . . . , n} satisfied det X | Y•,q det Y∼n,∼q = 0, then we would have n n+q det (X∼n,•) det Y = ∑ (−1) det X | Y•,q det Y∼n,∼q q=1 | {z } =0 (by Proposition 12.3, applied to i = n) = 0, which would contradict (46). Hence, not every q ∈ {1, 2, . . . , n} satisfies det X | Y•,q det Y∼n,∼q = 0. In other words, some q ∈ {1, 2, . . . , n} satisfies det X | Y•,q det Y∼n,∼q 6= 0. Consider this q. The q-th column of Y is the vector πn v f for some f ∈ B (by the definition of Y); consider this f . Thus, Y•,q = πn v f . Note that |B \ { f }| = |B| − 1 = n − 1 (since |B| = n). From det X | Y•,q det Y∼n,∼q 6= 0, we obtain det X | Y•,q 6= 0 and det Y∼n,∼q 6= 0. The (n − 1) × (n − 1)-matrix Y∼n,∼q is invertible (since det Y∼n,∼q 6= 0); thus, its n − 1 columns are linearly independent. But these n − 1 columns are precisely the The Bhargava greedoid as a Gaussian elimination greedoid page 55

n − 1 vectors πn−1 (ve) for all e ∈ B \ { f } (by the definition of Y and the choice of f ). Hence, the n − 1 vectors πn−1 (ve) for all e ∈ B \ { f } are linearly independent. In other words, the family (πn−1 (ve))e∈B\{ f } is linearly independent. In other words, the family π|B\{ f }| (ve) is linearly independent (since |B \ { f }| = n − 1). e∈B\{ f } In other words, B \ { f } ∈ G (by the definition of G). The n × n-matrix X | Y•,q is invertible (since det X | Y•,q 6= 0); thus, its n columns are linearly independent. But these n columns are precisely the n − 1 vectors πn (ve) for all e ∈ A as well as the extra vector Y•,q = πn v f . Thus, if we had f ∈ A, then these n columns would contain two equal vectors (namely, the extra vector Y•,q = πn v f would be equal to one of the vectors πn (ve) with e ∈ A), and thus would be linearly dependent. But this would contradict the fact that these n columns are linearly independent. Hence, we cannot have f ∈ A. Thus, f ∈/ A. Combining f ∈ B with f ∈/ A, we obtain f ∈ B \ A. From f ∈/ A, we also obtain |A ∪ { f }| = |A| + 1 = n (since |A| = n − 1). Recall that the n columns of the n × n-matrix X | Y•,q are the n − 1 vectors πn (ve) for all e ∈ A as well as the extra vector Y•,q = πn v f . In other words, they are the n vectors πn (ve) for all e ∈ A ∪ { f } (since f ∈/ A). Hence, the n vectors πn (ve) for all e ∈ A ∪ { f } are linearly independent (since the n columns of the n × n-matrix X | Y•,q are linearly independent). In other words, the family (πn (e))e∈A∪{ f } is linearly independent. In other words, the family π|A∪{ f }| (e) e∈A∪{ f } is linearly independent (since n = |A ∪ { f }|). In other words, A ∪ { f } ∈ G (by the definition of G). We now know that f ∈ B \ A and A ∪ { f } ∈ G and B \ { f } ∈ G. Hence, there exists b ∈ B \ A such that A ∪ {b} ∈ G and B \ {b} ∈ G (namely, b = f ). This shows that the axiom (iv) from Definition 12.1 is satisfied for F = G. Proof of the axiom (iii) for F = G: Clearly, axiom (iii) from Definition 12.1 follows from axiom (iv). Thus, axiom (iii) holds for F = G (since we already know that axiom (iv) holds for F = G). We have now showed that the system G satisfies the four axioms (i), (ii), (iii) and (iv) from Definition 12.1. Thus, G is a strong greedoid. This proves Theorem 1.6.

13. Appendix: Proof of Proposition 1.7

We shall now prove Proposition 1.7. This proof is mostly bookkeeping.

Proof of Proposition 1.7 (sketched). If Gk is empty, then the claim is obvious. Thus, we WLOG assume that Gk is nonempty. We assumed that G is a Gaussian elimination greedoid. In other words, there exist a ﬁeld K and a vector family (ve)e∈E over K such that G is the Gaussian elimination greedoid of this vector family (ve)e∈E. Consider these K and (ve)e∈E. k Let X be the k × |E|-matrix whose columns are the vectors πk (ve) ∈ K for all The Bhargava greedoid as a Gaussian elimination greedoid page 56 e ∈ E (in some order). Then, there is a bijection φ : {1, 2, . . . , |E|} → E such that the columns of the matrix X are πk vφ(1) , πk vφ(2) ,..., πk vφ(|E|) (in this order). Consider this φ. Hence, ( f -th column of X) = πk vφ( f ) (47) for each f ∈ {1, 2, . . . , |E|}. Let V be the vector matroid26 of the matrix X. This is a matroid on the set {1, 2, . . . , |E|}; its rank is at most k (since X has k rows and therefore rank 6 k). The bijection φ can be used to transport the matroid V onto the set E. More precisely, we can deﬁne a matroid V 0 on the set E by

V 0 = {φ (I) | I ∈ V}

(where we regard a matroid as a collection of independent sets). The independent sets of this matroid V 0 are the subsets I of E for which φ−1 (I) is an independent set of V. Moreover, the map φ is an isomorphism from the matroid V to the matroid V 0. Thus, the matroid V 0 is isomorphic to the vector matroid V, and hence is representable. Our goal is now to show that Gk is the collection of bases of V. Let F be a k-element subset of E. Then, we have the following chain of logical

26See [Oxley11, §1.1] for the deﬁnition of a vector matroid. The Bhargava greedoid as a Gaussian elimination greedoid page 57 equivalences:

(F ∈ Gk) ⇐⇒ (F ∈ G)(since Gk is the set of all k-element sets in G) F |F| ⇐⇒ the family π|F| (ve) ∈ K is linearly independent e∈F by the deﬁnition of a Gaussian elimination greedoid,

since G is the Gaussian elimination greedoid of (ve)e∈E F k ⇐⇒ the family (πk (ve))e∈F ∈ K is linearly independent (since |F| = k (because F is a k-element set)) φ−1(F) k ⇐⇒ the family πk vφ( f ) ∈ K is linearly independent f ∈φ−1(F)   since φ : {1, 2, . . . , |E|} → E is a bijection, and thus the   family πk vφ( f ) is just a reindexing of the family (πk (ve))e∈F f ∈φ−1(F) φ−1(F) k ⇐⇒ the family ( f -th column of X) f ∈φ−1(F) ∈ K is linearly independent since (47) yields ( f -th column of X) f ∈ −1(F) = πk vφ( f ) φ f ∈φ−1(F) ⇐⇒ φ−1 (F) is an independent set of the vector matroid of X (by the deﬁnition of a vector matroid) ⇐⇒ φ−1 (F) is an independent set of V (48) (since V is the vector matroid of X) . Forget that we ﬁxed F. We thus have proved the equivalence (48) for each k- element subset F of E. Recall that Gk is nonempty. In other words, there exists some B ∈ Gk. Consider this B. Thus, B is a k-element set in G (since Gk is the set of all k-element sets in G). Hence, the equivalence (48) (applied to F = B) yields the equivalence

−1 (B ∈ Gk) ⇐⇒ φ (B) is an independent set of V .

Thus, φ−1 (B) is an independent set of V (since B ∈ G ). Since φ is a bijection, k we have φ−1 (B) = |B| = k (since B is a k-element set). In other words, φ−1 (B) is a k-element set. Thus, the matroid V has a k-element independent set (namely, φ−1 (B)). Therefore, the rank of V is at least k. Since we also know that the rank of V is at most k, we thus conclude that the rank of V is exactly k. Hence, the rank of V 0 is exactly k as well (since the matroid V 0 is isomorphic to the matroid V). Therefore, the bases of V 0 are the k-element independent sets of V 0. Thus, a k-element set is an independent set of V 0 if and only if it is a basis of V 0. The Bhargava greedoid as a Gaussian elimination greedoid page 58

Now, let F again be a k-element subset of E. Thus, F is an independent set of V 0 if and only if F is a basis of V 0 (since a k-element set is an independent set of V 0 if and only if it is a basis of V 0). Also, φ−1 (F) is an independent set of V if and only if F is an independent set of V 0 (since the independent sets of the matroid V 0 are the subsets I of E for which φ−1 (I) is an independent set of V). Now, we have the following chain of logical equivalences:

−1 (F ∈ Gk) ⇐⇒ φ (F) is an independent set of V (by (48)) ⇐⇒ F is an independent set of V 0 ! since φ−1 (F) is an independent set of V if and only if F is an independent set of V 0 ⇐⇒ F is a basis of V 0

(since F is an independent set of V 0 if and only if F is a basis of V 0). Forget that we ﬁxed F. We thus have proved that, for each k-element subset F of E, we have the equivalence

0 (F ∈ Gk) ⇐⇒ F is a basis of V .

0 0 Hence, Gk is the collection of all bases of V (since all bases of V are k-element 0 subsets of E (because V is a matroid of rank k)). Hence, Gk is the collection of bases of a representable matroid on the ground set E (since V 0 is a representable matroid on the ground set E). This proves Proposition 1.7.

14. Appendix: Proofs of some properties of L

We shall next prove four lemmas that were left unproven in Section 6: namely, Lemma 6.4, Lemma 6.8, Lemma 6.11 and Lemma 6.12. Proof of Lemma 6.4. (a) Let a ∈ L be a nonzero element. We must prove that the element a belongs to L+ if and only if its order ord a is nonnegative (i.e., we have ord a > 0). In other words, we must prove that a ∈ L+ if and only if ord a > 0. Recall that ord a is deﬁned as the smallest β ∈ V such that tβ a 6= 0. Thus, ord a is a β ∈ V such that tβ a 6= 0. In other words, ord a ∈ V and [tord a] a 6= 0. We shall now prove the implication

(a ∈ L+) =⇒ (ord a > 0) . (49)

[Proof of (49): Assume that a ∈ L+. Then, a ∈ L+ = K [V>0]. In other words, a is a K-linear combination of the family (tα) ∈V . Thus, every α ∈ V that satisﬁes α >0 [tα] a 6= 0 is an element of V>0. Applying this to α = ord a, we conclude that ord a is an element of V>0 (since [tord a] a 6= 0). In other words, ord a > 0. Thus, the implication (49) is proved.] The Bhargava greedoid as a Gaussian elimination greedoid page 59

Next, let us prove the implication (ord a > 0) =⇒ (a ∈ L+) . (50) [Proof of (50): Assume that ord a > 0. Now, let α ∈ V be such that α < 0. We shall show that [tα] a = 0. Indeed, assume the contrary. Thus, [tα] a 6= 0. But recall that ord a is the smallest β ∈ V such that tβ a 6= 0. Hence, for any β ∈ V satisfying tβ a 6= 0, we must have β > ord a. Applying this to β = α, we obtain α > ord a > 0. This contradicts α < 0. This contradiction shows that our assumption was wrong. Hence, we have proved that [tα] a = 0. Forget that we ﬁxed α. We thus have proved that

[tα] a = 0 for each α ∈ V satisfying α < 0. (51) But a ∈ L = K [V]. Thus,

a = ∑ ([tα] a) · tα = ∑ ([tα] a) · tα + ∑ ([tα] a) ·tα α∈V α∈V; α∈V; | {z } = α>0 α<0 0 |{z} (by (51)) = ∑ α∈V >0 (since each α ∈ V satisﬁes either α > 0 or α < 0 (but not both)) = ∑ ([tα] a) · tα + ∑ 0tα = ∑ ([tα] a) · tα. ∈V ∈V ∈V α >0 α ; α >0 α<0 | {z } =0

Thus, a is a K-linear combination of the family (tα) ∈V . In other words, a ∈ α >0 K [V>0]. In other words, a ∈ L+ (since L+ = K [V>0]). This proves the implication (50).] Combining the two implications (49) and (50), we obtain the equivalence (a ∈ L+) ⇐⇒ (ord a > 0) . In other words, a ∈ L+ if and only if ord a > 0. This proves Lemma 6.4 (a). (b) Let a ∈ L be nonzero. The deﬁnition of ord a shows that ord a = the smallest β ∈ V such that tβ a 6= 0 . The same argument (applied to −a instead of a) yields     ord (−a) = the smallest β ∈ V such that tβ (−a) 6= 0  | {z }  =−[tβ]a = the smallest β ∈ V such that − tβ a 6= 0 = the smallest β ∈ V such that tβ a 6= 0 since the condition “ − tβ a 6= 0” is equivalent to “ tβ a 6= 0” . The Bhargava greedoid as a Gaussian elimination greedoid page 60

Comparing these two equalities, we obtain ord (−a) = ord a. This proves Lemma 6.4 (b). (c) This is analogous to the standard fact that any two nonzero polynomials f , g ∈ K [X] satisfy f g 6= 0 and deg ( f g) = deg f + deg g. For the sake of completeness, let us nevertheless present the proof. Recall that ord a is defined as the smallest β ∈ V such that tβ a 6= 0. Thus, ord a is a β ∈ V such that tβ a 6= 0. In other words, ord a ∈ V and [tord a] a 6= 0. The same argument (applied to b instead of a) yields ord b ∈ V and [tord b] b 6= 0. From [tord a] a 6= 0 and [tord b] b 6= 0, we obtain ([tord a] a) · ([tord b] b) 6= 0 (since [tord a] a and [tord b] b belong to the integral domain K). Moreover, tβ a = 0 for each β ∈ V satisfying β < ord a (52) (since ord a is the smallest β ∈ V such that tβ a 6= 0). Similarly, tβ b = 0 for each β ∈ V satisfying β < ord b. (53) For any α ∈ V, we have by the definition of the multiplication [t ](ab) = t a · t b α ∑ β α−β in the group algebra L = K [V] β∈V = ∑ tβ a · tα−β b + ∑ tβ a · tα−β b β∈V; | {z } β∈V; β (since each β ∈ V satisfies either β < ord a or β > ord a (but not both)) = ∑ 0 · tα−β b + ∑ tβ a · tα−β b β∈V; β∈V; βord a | {z } =0 = ∑ tβ a · tα−β b . (54) β∈V; β>ord a Now, let α ∈ V be such that α < ord a + ord b. Then, for every β ∈ V satisfying β > ord a, we have α − β < (ord a + ord b) − ord a = ord b |{z} ord a and thus tα−β b = 0 (55) (by (53), applied to α − β instead of β). Hence, (54) becomes [tα](ab) = ∑ tβ a · tα−β b = ∑ tβ a · 0 = 0. β∈V; | {z } β∈V; β ord a =0 β ord a > (by (55)) > The Bhargava greedoid as a Gaussian elimination greedoid page 61

Forget that we ﬁxed α. We thus have proved that

[tα](ab) = 0 for any α ∈ V satisfying α < ord a + ord b. (56)

On the other hand, for every β ∈ V satisfying β > ord a, we have

(ord a + ord b) − β < (ord a + ord b) − ord a = ord b |{z} >ord a and thus h i t(ord a+ord b)−β b = 0 (57) (by (53), applied to (ord a + ord b) − β instead of β). Hence, (54) (applied to α = ord a + ord b) yields

[tord a+ord b](ab) h i = ∑ tβ a · t(ord a+ord b)−β b β∈V; β>ord a h i h i = ([tord a] a) · t(ord a+ord b)−ord a b + ∑ tβ a · t(ord a+ord b)−β b | {z } β∈V; | {z } =[tord b]b β>ord a =0 (since (ord a+ord b)−ord a=ord b) (by (57)) (here, we have split off the addend for β = ord a from the sum) = ([tord a] a) · ([tord b] b) + ∑ tβ a · 0 = ([tord a] a) · ([tord b] b) 6= 0. β∈V; β>ord a | {z } =0 Thus, ord a + ord b is a β ∈ V such that tβ (ab) 6= 0. In view of (56), we conclude that ord a + ord b is the smallest such β. From [t + ](ab) 6= 0, we obtain ab 6= 0. Thus, ab is a nonzero element of ord a ord b L. Hence, ord (ab) is defined as the smallest β ∈ V such that tβ (ab) 6= 0. But we already know that ord a + ord b is the smallest such β. Comparing these two results, we conclude that ord (ab) = ord a + ord b. This proves Lemma 6.4 (c). (d) Assume the contrary. Thus, ord (a + b) < min {ord a, ord b} 6 ord a. Simi- larly, ord (a + b) < ord b. Recall that ord a is defined as the smallest β ∈ V such that tβ a 6= 0. Hence, in particular, no such β is smaller than ord a. In other words, tβ a = 0 for each β ∈ V satisfying β < ord a. h i h i Applying this to β = ord (a + b), we find tord(a+b) a = 0. Similarly, tord(a+b) b = 0. The Bhargava greedoid as a Gaussian elimination greedoid page 62

But ord (a + b) is defined as the smallest β ∈ V such that tβ (a + b) 6= 0. Thus, ord (a + b) is a β ∈ V such that tβ (a + b) 6= 0. In other words, ord (a + b) ∈ V h i h i and tord(a+b) (a + b) 6= 0. But tord(a+b) (a + b) 6= 0 contradicts h i h i h i tord(a+b) (a + b) = tord(a+b) a + tord(a+b) b = 0 + 0 = 0. | {z } | {z } =0 =0 This contradiction shows that our assumption was wrong. This proves Lemma 6.4 (d). Proof of Lemma 6.8. First, let us observe that the distance function d : E × E → V in Definition 6.7 is well-defined. [Proof: We must show that − ord (a − b) ∈ V is well-defined for each (a, b) ∈ E × E. So let us fix (a, b) ∈ E × E. Thus, a and b are two distinct elements of E (by the definition of E × E). Since a and b are distinct, we have a − b 6= 0. Hence, the element a − b of L is nonzero, and therefore ord (a − b) ∈ V is well-defined. Thus, − ord (a − b) ∈ V is well-defined. Forget that we fixed (a, b). We thus have showed that − ord (a − b) ∈ V is well- defined for each (a, b) ∈ E × E. This completes our proof.] Now, consider the distance function d : E × E → V in Definition 6.7. Our goal is to prove that (E, w, d) is a V-ultra triple whenever w : E → V is a function. According to the definition of a “V-ultra triple”, this boils down to proving the following two statements:

Statement 1: We have d (a, b) = d (b, a) for any two distinct elements a and b of E.

Statement 2: We have d (a, b) 6 max {d (a, c) , d (b, c)} for any three distinct elements a, b and c of E.

Let us prove these two statements. [Proof of Statement 1: Let a and b be two distinct elements of E. Thus, a − b 6= 0 (since a and b are distinct), so that a − b is a nonzero element of L. Hence, Lemma 6.4 (b) (applied to a − b instead of a) yields ord (− (a − b)) = ord (a − b). But   the deﬁnition of d yields d (a, b) = ord (a − b) and d (b, a) = ord  b − a  =  | {z }  =−(a−b) ord (− (a − b)) = ord (a − b). Comparing these two equalities, we ﬁnd d (a, b) = d (b, a). This proves Statement 1.] [Proof of Statement 2: Let a, b and c be three distinct elements of E. Then, State- ment 1 (applied to c instead of a) yields d (c, b) = d (b, c). We have a − b 6= 0 (since a and b are distinct), so that a − b is a nonzero element of L. Likewise, a − c and c − b are nonzero elements of L. Now, the nonzero The Bhargava greedoid as a Gaussian elimination greedoid page 63 elements a − c and c − b of L have the property that their sum (a − c) + (c − b) is nonzero (since (a − c) + (c − b) = a − b is nonzero). Thus, Lemma 6.4 (d) (applied to a − c and c − b instead of a and b) yields ord ((a − c) + (c − b)) > min {ord (a − c) , ord (c − b)}. In view of (a − c) + (c − b) = a − b, this rewrites as

ord (a − b) > min {ord (a − c) , ord (c − b)} .

Hence,

− ord (a − b) 6 − min {ord (a − c) , ord (c − b)} | {z } >min{ord(a−c),ord(c−b)} = max {− ord (a − c) , − ord (c − b)} . (58)

But the deﬁnition of d yields

d (a, b) = − ord (a − b) and d (a, c) = − ord (a − c) and d (c, b) = − ord (c − b) .

In view of these equalities, we can rewrite (58) as d (a, b) 6 max {d (a, c) , d (c, b)}. In view of d (c, b) = d (b, c), this further rewrites as d (a, b) 6 max {d (a, c) , d (b, c)}. This proves Statement 2.] We thus have proved both Statements 1 and 2. Therefore, (E, w, d) is a V-ultra triple whenever w : E → V is a function (by the deﬁnition of a “V-ultra triple”). This proves Lemma 6.8.

Proof of Lemma 6.11. The unity of the K-algebra L+ = K [V ] is 1L = t . Hence, >0 + 0 the map π sends this unity to π 1L+ = π (t0) = [t0] t0 = 1. Also, the map π is clearly K-linear (since any x, y ∈ L+ satisfy [t0](x + y) = ([t0] x) + ([t0] y), and since any x ∈ L+ and λ ∈ K satisfy [t0](λx) = λ [t0] x). Let x, y ∈ L+. We shall prove that π (xy) = π (x) · π (y). We have x ∈ L+ = K [V>0]. In other words, x is a K-linear combination of the family (tα)α∈V . In other words, we can write x in the form x = ∑ λαtα for >0 ∈V α >0 V 0 some family (λα) ∈V ∈ K > of coefficients λα (such that all but finitely many α >0 α ∈ V>0 satisfy λα = 0). Consider this family (λα)α∈V . From x = ∑ λαtα, we >0 ∈V α >0 obtain [t0] x = λ0. Thus, the definition of π yields

π (x) = [t0] x = λ0. (59)

We have y ∈ L+ = K [V>0]. In other words, y is a K-linear combination of = the family (tα)α∈V tβ ∈V . In other words, we can write y in the form >0 β >0 V y = t ∈ K >0 ∑ µβ β for some family µβ β∈V of coefﬁcients µβ (such that all ∈V >0 β >0 The Bhargava greedoid as a Gaussian elimination greedoid page 64

∈ V = but ﬁnitely many β >0 satisfy µβ 0). Consider this family µβ ∈V . From β >0 y = ∑ µβtβ, we obtain [t0] y = µ0. Thus, the deﬁnition of π yields ∈V β >0

π (y) = [t0] y = µ0. (60)

For any pair (α, β) ∈ V>0 × V>0 satisfying (α, β) 6= (0, 0), we have [t0] tα+β = 0 (61)

27. Now, multiplying the equalities x = ∑ λαtα and y = ∑ µβtβ, we obtain ∈V ∈V α >0 β >0    

xy =  ∑ λαtα  ∑ µβtβ = ∑ ∑ λαµβ tαtβ ∈V ∈V ∈V ∈V α >0 β >0 α >0 β >0 |{z} | {z } =tα+β = ∑ (α,β)∈V ×V >0 >0 = ∑ λαµβtα+β. ( )∈V ×V α,β >0 >0 Applying the map π to both sides of this equality, we obtain    

π (xy) = π  ∑ λαµβtα+β = [t0]  ∑ λαµβtα+β ( )∈V ×V ( )∈V ×V α,β >0 >0 α,β >0 >0 (by the deﬁnition of π) = ∑ λαµβ [t0] tα+β ( )∈V ×V α,β >0 >0 = λ0 µ0 [t0](t0+0) + ∑ λαµβ [t0] tα+β |{z} |{z} | {z } (α,β)∈V ×V ; | {z } =π(x) =π(y) =[t ](t ) >0 >0 = 0 0 (α,β)6=(0,0) 0 (by (59)) (by (60)) =1 (by (61)) (here, we have split off the addend for (α, β) = (0, 0) from the sum) = π (x) π (y) + ∑ λαµβ0 = π (x) π (y) . ( )∈V ×V α,β >0 >0; (α,β)6=(0,0) | {z } =0

27 Proof of (61): Let (α, β) ∈ V>0 × V>0 be a pair satisfying (α, β) 6= (0, 0). From (α, β) ∈ V>0 × V>0, we obtain α ∈ V>0. In other words, α > 0 (by the deﬁnition of V>0). Similarly, β > 0. At least one of the two elements α and β is 6= 0 (since (α, β) 6= (0, 0)). In other words, we have α 6= 0 or β 6= 0. Since α and β play symmetric roles in our setting, we can WLOG assume that α 6= 0 (since otherwise, we can achieve this by swapping α with β). Combining α 6= 0 with α > 0, we obtain α > 0. Adding this inequality to β > 0, we obtain α + β > 0 + 0 = 0. Hence, α + β 6= 0, so that [t0] tα+β = 0. This proves (61). The Bhargava greedoid as a Gaussian elimination greedoid page 65

Forget that we ﬁxed x and y. We thus have showed that π (xy) = π (x) · π (y) for all x, y ∈ L+. Hence, π is a K-algebra homomorphism (since π is K-linear and π 1L+ = 1). This proves Lemma 6.11.

Proof of Lemma 6.12. Lemma 6.4 (a) shows that a belongs to L+ if and only if its order ord a is nonnegative. Hence, ord a is nonnegative (since a belongs to L+). In other words, ord a > 0. Recall that ord a is deﬁned as the smallest β ∈ V such that tβ a 6= 0. Thus, ord a is a β ∈ V such that tβ a 6= 0. In other words, ord a ∈ V and [tord a] a 6= 0. We shall ﬁrst prove the implication

(ord a = 0) =⇒ (π (a) 6= 0) . (62)

[Proof of (62): Assume that ord a = 0. Then, 0 = ord a, so that [t0] a = [tord a] a 6= 0. But the deﬁnition of π yields π (a) = [t0] a 6= 0. Thus, the implication (62) is proved.] Next, let us prove the implication

(π (a) 6= 0) =⇒ (ord a = 0) . (63)

[Proof of (63): Assume that π (a) 6= 0. But the definition of π yields π (a) = [t0] a. Hence, [t ] a = π (a) 6= 0. 0 Now, recall that ord a is the smallest β ∈ V such that tβ a 6= 0. Hence, if β ∈ V satisfies tβ a 6= 0, then β > ord a. Applying this to β = 0, we obtain 0 > ord a (since [t0] a 6= 0). Combining this with ord a > 0, we find ord a = 0. This proves the implication (63).] Combining the two implications (63) and (62), we obtain the logical equivalence (π (a) 6= 0) ⇐⇒ (ord a = 0). In other words, π (a) 6= 0 holds if and only if ord a = 0. This proves Lemma 6.12.

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